Mathematics 2 DEMO - Scope and Sequence and 4 sample units by Grupo Anaya, S.A.

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s c i t a m e h t a M Albero u l e t z Ga nez, I. é m i J s ra a Caña r e J. Cole l o C R.

Building

Blocks

this is

your book

Reading and listening

CH UNIT

F EA THE OPENING PAGES O

We read and listen a brief historical introduction of the contents your are going to learn in the unit.

You can do these motivating activities to activate your previous knowledge.

Pythagorean theore

Pythagorean theorem is right-angled: When the blue triangle A=B+C

PYTHAGOREAN THEOREM

square is equal to the The area of the large two small ones. sum of the areas of the

Reading and listenin

3 What

Babylonians Egyptians and ical concept. As you is an important geometr of the sides of any the squares The Pythagorean theorem relationship between know, it describes the knew right-angled triangle. s and the Babylonians ago, both the Egyptian them to construct used they More than 3 000 years sides were right-angled and , used triangles whose that certain triangles Egyptians, for example sacred) to divide fields and right angles. The ed 5 (which they consider and 4 3, d measure build pyramids.

cm2? 2 = 9 cm and C = 16 is the area of A if B

icular to Euclid’s proof drew a line perpend rean theorem, Euclid large square To prove the Pythago line), dividing the the triangle (the green and C = A2. the longer side of = A1 He showed that: B into two rectangles. A1 A2

B C

Egypt and Pythagoras th BCE) travelled to ras (6 century es. His great In his youth, Pythago about these properti g the undoubtedly learnt Babylon, where he general theorem describin gled come up with the right-an achievement was to the sides of any squares drawn on relationship between his name. this theorem bears why is That triangle.

and answer. 4 Look at the figure the grid as Taking a square from A1 a unit: A2 squares does a) How many unit 9 B, contain? the small square, do you What ? A 1 rectangle 16 And B 15 notice? number of unit 20 b) Check that the in as same the squares in C is A2. C . c) Describe this property it confirm? d) What theorem does State the theorem. cm. Write 5 cm, 12 cm and 13 of sides with gled triangle the lengths of the sides. 5 Draw a right-an g the relationship between an equality expressin

ts Euclid, Euclid's Elemen the theorem, but Pythagoras who proved Elements around the year However, it was not ria wrote d and Euclid of Alexand he collected, organise g it two centuries later. of 13 books in which ge of his time, providin 300 BCE. It is a set the mathematical knowleddemonstrated what we now expanded upon all I he Book In . structure with a solid logical ean theorem. Pythagor the as know verb ‘come up with’. to define the phrasal verb in 1 Use context clues k using this phrasal sentence in your noteboo Then, write a new . the a different situation write sentences about ending and past continuous, with Pythagoras and 2 Using past simple starting rean theorem history of the Pythago with Euclid.

LANG

178

CONTENT DEVELOPMENT AND ACTIVITIES 3 APPLICATIONS OF THE PYTHAGOREAN THEOREM

PYTHAGOREAN THEOREM

Consolidating

The two shortest sides of a right-angled triangle form a right angle. They are called the legs. The longest side is called the hypotenuse. In general, we say that a is the hypotenuse and b and c are the legs. a

b c

This observation was made by the Chinese 400 years before Pythagoras was born. anayaeducacion.es GeoGebra. Graphical demonstration of the Pythagorean theorem.

Look at this demonstration:

If we compare both figures, it is clear that a 2 = b 2 + c 2.

a c

What are the areas of the unknown squares in the following shapes?

102 m2

2 = reduce 1.2 m the length of the Looking at the blue triangle: AC = 1.2 2•+If…we … m the size of the right angle making it acute, Therefore: 2 = … mside decreases, as does the area of the corresponding B square. Looking at the red triangle: d = … 2 + 4.8opposite 1.6 m A 2 2 2 2 2 2 The diagonal of the cuboid measures 5.2 m.Figure C 8 4.5 is less than 3 + 4 8 In general: a <Db + c 4.8 m ideas Note that you can calculate it directly: idating Consol

d = 1.2 2 + 1.6 2 + 4.8 2 = f = 5.2 m

In general, in a cuboid with dimensions a Ò b Ò c the diagonal is: a) b) d = a 2 + b 2 + c 2 A1

57 cm2

87 m2

B 1.6 m

1.2 m A 3 dm2

13 Calculate the height, h, of a regular pyramid that has a square base with 30 cm

sides, and a lateral face with an area of 255 m2. cm2 the height, h, of the lateral face. First, we 57 find 31 m2 A 255 = 30 · a 8 510 = 30a 8 a = … m A1 = 57 + 57 = … cm2 2 A2 = 87 + … = … m2 The height of the lateral face is the hypotenuse of the triangle shown on the right:

8, 15, 17

12, 35, 37

5, 12, 13

9, 40, 41

13, 84, 85

7, 24, 25

3, 4, 5

11, 60, 61

16, 63, 65

Notice that if c, b, a is a Pythagorean triple, then kc, kb and ka are too. For example, 6, 8, 10 (the result of multiplying each of the components of the triple 3, 4, 5 by 2) is a Pythagorean triple.

70 2 + 240 2 = 4 900 + 57 600 = 62 500

Let’s practise! 245 2 = 60 025

of 54 cm. 15 2 + 36 2 = 225 + 1296 = f

d) 15 km, 20 km, 25 km a

Examples and solved problems. To put into practice the most important methods.

Find the diagonal of a square with a perimeter of 28 dam.

The parallel sides of a right trapezium are 13 dm and 19 dm long, and the oblique side is 10 dm. Calculate the height.

Calculate the identical sides of an isosceles triangle, knowing that the non-identical side is 5 m long, and the corresponding height is 6 m.

Calculate the length of the side of a rhombus with diagonals of 1 dm and 2.4 dm.

Find the height of an equilateral triangle with 40 cm sides. Round to the nearest millimetre.

Find the apothem of a regular hexagon with 20 cm sides. Remember that in a regular hexagon the side and the radius have the same length.

A regular pentagon with 11.7 cm sides is inscribed in a circumference with a radius of 10 cm. Calculate the apothem.

A straight line passes 10 cm from the centre of a circumference that has a radius of 15 cm. Find the length of the resulting chord, rounding your answer to the tenths.

How far from the centre of a circumference with an 8 cm radius must a line pass so that the chord measures 8 cm?

Calculate the diagonal of a cube with 20 cm sides. Round to the nearest millimetre.

f ) 21 mm, 42 mm, 21 mm g) 18 cm, 80 cm 82 cm

15 m

a = … dm2 A3 = 14h – …

Calculate the unknown side in each right-angled triangle: a)

b) 65 mm

15 m 20 m

4 A 13-metre chord is drawn on a circumference with

a radius of 9.7 m. How far is the line from its centre? 4 8 392 … 152 + 362 8 The triangle is… 2 = 1521 2 Find the and perimeter of an isosceles trapezium 5 A regular pentagon is inscribed in a circumference 39area with 3.2 m and 6.4 m bases, and 6.3 m height. with a radius of 1 m. Its perimeter is 5.85 m. Calculate c) 18 m, 80 m, 83 m the area. 3 Calculate the area of a regular hexagon with 18 cm + 6a400 = 6 724 18 2 + 80 2 = that 324 in length of the sides. (Remember regular hexagon 2 8the 4 8 the 832side … 182 +680Find The triangle is…diagonal of a cuboid with sides 2=f of 8 dm, 6 dm and 14 dm. and the83 radius have the same length.)

186

Consolidating Ideas. Exercises to co mplete and consolidat e the theory that the teacher has explained to you.

e) 17 miles, 10 miles, 5 miles

16 mm

Calculate the unknown side of each triangle and round it off to the tenths. a)

16 m 12 cm

anayaeducacion.es GeoGebra. Calculating areas applying the Pythagorean theorem. 180

12 dm

b) 35 m, 12 m, 37 m

14 dm2

Calculate the perimeter of a rectangle with a diagonal of 5.8 cm, and one side that measures 4 cm.

Say whether each of the following triangles is right-angled, acute-angled or obtuse-angled:

c) 23 dm, 30 dm, 21 dm

4 8 2452 < 702 + 2402 8 The triangle is…

15area dm,of36 39 dm triangle with a perimeter 1 Findb)the andm, equilateral

4 cm

a) 15 cm, 10 cm, 11 cm A

17 cm

21 dm

Calculate the area of the following squares:

2 – …2 = 2 Copy and complete to find each 30 m h = out …whether fof=the 8 mfollowing triangles is right-angled, acute or obtuse: 30 m Thea)pyramid 8m high. 70 cm, is 240 cm, 245 cm

ã Pythagorean triples If three natural numbers, c, b, a, satisfy* c 2 + b 2 = a 2, in other words, if they could be the measurements of the sides of a right-angled triangle, we say that the numbers form a Pythagorean triple. Here are a few of them:

F ocus on English

30 cm2

1 Copy these shapes in your notebook. Draw the square that is missing in each one and say what4.8the m area is. C

S1 386 dm2

acute triangle

If wecm expand the right angle making it obtuse, the length of the opposite side r = f =• 7.07 increases, as does the area of the corresponding square. Therefore:

60 m2

14 cm2

10 cm

1 cm

45 m2

Find the perimeter of the following shape:

Calculate the area of the green square in each of the following cases: a)

D 2 is greater 12 Find the diagonal of a cuboid with dimensions of B 1.28m,61.6 m and 4.8 thanm.32 + 42 8 In general: a 2 >C b 2 + c 2 Figure

As both triangles are right-angled, in both cases the area of the biggest square is equal to the sum of the areas of the smaller squares. Therefore: S2 = 386 dm2 – 47 dm2 = 339 dm2 S1 = 183 m2 + 102 m2 = 285 m2

satisfy: to fulfill a condition.

23 cm

• We already the relationship between the areas of the squares built on the 10 cm. Remember that the diagonals of the square are know perpendicular. r r sidesand of ahas right-angled triangle: Therefore, the coloured triangle is right-angled legs of equal length. 2 is equal to 32 + 42 8 In general: a 2 = b 2 + c 2 =… A 88r 25= … r 2 + r 2 = …2 8 2r 2Figure

Problem solved

47 dm2

18 cm

r O

… ≈ 643.7 •A Ifcircle a2 < = b 2πr + c 2=, … the ·triangle is an cm acute triangle. right-angled triangle obtuse triangle 11 Calculate the radius of the circumscribed circumference of this square with sides of

Unit 9

Pythagorean theorem

Therefore, the triangle PTO has a right angle at T :

unit

from this resources to choose Remember io. for your portfol

• If a 2 > b 2 + c 2, the 2 is an 2 2 triangle 2 2 obtuse triangle.r = OP – PT = … – … = …

183 m2

The icons included with some activities indicate the keys to the project.

10Remember The distance of a point P to the centre O of aAcircumference is OP B= 23 cm. We If draw we know the sidesfrom of a P triangle, a tangent to thewe circumference. The PT tangent segment is 18 cm. can findthe outarea whether it iscircle. right-angled: Find of the

The two identical squares have b + c as sides.

Look at the diagrams and analyse the explanation below:

• The If a 2tangent = b 2 + c 2line , the is is triangle perpendicular to the radius. right-angled.

According to the Pythagorean theorem: a 2 = b 2 + c 2 This means that the area of the square built on the hypotenuse is equal to the sum of the areas of the squares built on the legs. This relationship is only true if the triangle is right-angled.

a2 = b2 + c2 Interesting fact!

Unit 9

ã The sides of a triangle determine its type

ideas

12 cm

divided Each unit is phs and into epigra hs. subepigrap portant The most im in bold. re a contents

Focus on English. Do you think Mathematics and English haveanything PROBLEMS EXERCISES AND in common? Discover how language and mathematics are linked so you can learn both: Mathematics and English.

17 m

32 mm

28 mm

The audios of each unit’s content are available at www.anayaeducacion.es

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Find the diagonal of a cuboid with sides of 3 cm, 4 cm and 12 cm.

A 24 m tangent segment is drawn from an external point P to a circumference with a 10 m radius. How far is P from the centre of the circumference?

181

187

Let’s practise! These are exercises to apply the theory you have learnt.

KEYS

PROJECT

SDG SDG Commitment Discover the Sustainable Development Goals and be an active part of our commitment to make a more equal and liveable world.

Developing thinking Work on strategies for thinking: reflect on the content you are learning, generate ideas, organise them, debate them, explain them…

Cooperative learning Get involved in your learning and participate in the group’s learning; you will find that cooperating improves performance and harmony in the class.

Emotional education Get to know yourself; identify the situations that bring up complicated emotions and manage them with constructive, self-affirming experiences.

EXERCISES AND PROBLEMS 3 APPLICATIONS OF THE PYTHAGOREAN THEOREM

from this resources to choose Remember io. for your portfol

unit

Unit 9

PROBLEMS EXERCISES PROBLEMS S AND EXERCISEAND

idating ideas

Consol

18 cm

23 cm

geometrical shapes. Round to one decimal place. Calculate the area of the green square in each of a)the following cases: b) a) b)

13 Calculate the height, h, of a regular pyramid that has a square base with 30 cm

sides, and a lateral face with an area of 255 m2. First, we find the height, h, of the lateral face. 255 = 30 · a 8 510 = 30a 8 a = … m 2 The height of the lateral face is the hypotenuse of the triangle shown on the right: h = …2 – …2 = f = 8 m

Glossary A

associative property billón

The pyramid is 8 m high.

additive system

5.6are dm grouped. g) 18not cm,depend 80 cm on 82 how cm the sumands c) A 1 followed by 12 zeros d) (One million millions). Calculate the unknown side in each right-angled To join several quantities (addends) into one. triangle:

commutative property

Let’s practise!

decimal numeral system

1 Find the area of an equilateral triangle with a perimeter

4 A 13-metre chord is drawn on a circumference with

2 Find the area and perimeter of an isosceles trapezium

5 A regular pentagon is inscribed in a circumference

with 3.2 m and 6.4 m bases, and 6.3 m height.

3 Calculate the area of a regular hexagon with 18 cm

sides. (Remember that in a regular hexagon the side and the radius have the same length.)

a radius of 9.7 m. How far is the line from its centre? distributive property 5

186

6 Find the length of the diagonal of a cuboid with sides integer division

of 8 dm, 6 dm and 14 dm.

million

multiplication anayaeducacion.es GeoGebra. Calculating areas applying the Pythagorean theorem. natural numbers

10 m

8 dm

3 dm

d 10

Find the height of an equilateral triangle with 40 cm sides. Round to the nearest millimetre.

13d)

Find the apothem of a regular hexagon with 20 cm 5m sides. Remember5 mthat in a regular hexagon the side and the radius have the same length.

15 23

Problem solving

Is this triangle right-angled?

Calculate the length 4 dm of the side of a rhombus with diagonalsmof 1 dm and 2.4 dm.

Problem solved 28

20 cm

13 cm

A 14.5 m-high electricity pole breaks at the base and falls against a building located 10 m away from it. At what height does the pole hit the building?

5 cm

Solution: We first calculate the unknown side CB . A To express a number as a product of its divisors. 20 13 of athe common divisors of two or more numbers. The greatest

divisor (GCD)

A straight line passes 10 cm from the lowest centre of common a circumference that has a radius of 15 cm. multiple Find the(LCM) length of the resulting chord, rounding your answer to multiple Problem the tenths. solved

C of the common multiples B of two or more numbers. The lowest 5 M x 2

During a carnival in my town, we hang

a = 13 – 5 = 144 8 a = 144 8 a = 12 a 1 m-high piñata in the middle of a 34 m-long rope, Number containing another number an exact number of times. which is tied to two 12 m-high poles that are 30 m apart. How far from the ground is the piñata?

2 = 202 – a 2 = 256 8 x = 256 8 x = 16 xNumber that is only divisible by itself and the unit.

CB = 5 + x = 21 cm 20

17 Calculate the diagonal of a cube with 20 cm sides. Solution: 4. INTEGERS Calculate side that of each A propertythe of unknown multiplication saystriangle that theand product of the not 2 +multiplication 2 =millimetre. x 2 =to42the8nearest 2xdoes 16 xRound absolute value The natural number21we get from removing its sign. round it off to the2 remove tenths. the brackets. change if we km 2 integer 8 the x = diagonal 8 ≈ 2.83of a cuboid with sidesof ofan 3 cm, a) The distribution b) of a whole among c) several, equal parts.18 x = 8Find 13 2 + 20 2 = 569 4 8 212 < 132 + 202 4 cm and 12 2cm. opposite of an integer Another A = x $ x = x = 8 = 4 mm2 21 2 = integer 441 with the same absolute value, but with the opposite sign. Division where the remainder is zero. 2 2 2 32 mm x set The set of all positive natural numbers, zero, and the negatives of the natural 16 m 17 m 19 A 24 m tangent segment is drawn from anZ external Division where the remainder is not zero. It is an acute triangle. P = x + x + 4 = 2 · 2.83 + 4 = 9.66 mm numbers. point P to a circumference with a 10 m radius. How 30 m A 1 followed by 6 zeros. 12 cm far is P from the centre of the circumference?

division with a radius of 1 m. Its perimeter is 5.85 m. Calculate the area. exact division

We learn the relevant terms that are underlined in the units with a clear definition.

39 m

16 m

Numeral system where adding symbols adds their represented amount. prime number b) 65 mm 22 cm x 9.6property cm Calculate theof area and perimeter mmsays that 16the result How far from thedoes centre of a circumference with A of addtion and multiplication 16 that the sum 15 m of the that coincides with not changexif the order of the sumands changes. an triangle 8 cm radius must a line pass so that the chord 10 m half of a square whose diagonal Positional numeral system with ten symbols or figures (0,measures 1, 2, 3, 4, 8 cm? 5, 6, 7, 8 and 9). x 20 m measures 4 mm. 4 mm This is the numeral system we currently use.

12 cm

of 54 cm.

25 mm

addition

The exercises are divided into topics. Each one is also marked with its degree of difficulty, from one to three.

9 mm

13 m

inscribed 5Amregular pentagon with 11.7 cm sides is factorise in a circumference with a radius of 10 cm. Calculate greatest common the5 result apothem. f ) 21Amm, 42 mm, 21 mm and multiplication that says that the m property of addition of the sum does

15 m

30 m

12 dm

Areas and perimeters using the 3 Pythagorean Say whether theorem each of the following triangles is 21 right-angled, Find the area and perimeter of these shapes. To do acute-angled or obtuse-angled: so, will 10 firstcm, have a) you 15 cm, 11to cmcalculate the unknown length of one of their elements. If they are not exact, find b) 35tom, 12decimal m, 37 m them one place. c) 23 dm, 30 dm, 21 dm 2.4 dm b) a) d) 15 km, 20 km, 25 km

25 mm 1. NATURAL NUMBERS e) 17 miles, 10 miles, 5 miles

h h

30 m

20 d

m The20parallel sides of a right trapezium are 13 dm and 19 dm long, and the oblique side is 10 dm. Calculate the height.

3m Calculate the identical sides of an isosceles triangle, knowing that the non-identical side is 5 m long, and c) the corresponding height is 6 m.

21 dm

1.2 m 1.6 m

Find the diagonal of a square with a perimeter of 28 dam.

1.6 m

4.8 m

C B

d = a 2 + b 2 + c 2

x 36 m

3 dm

mm 20

28 mm

d = 1.2 2 + 1.6 2 + 4.8 2 = f = 5.2 m

4 cm

20 cm

4.8 m

In general, in a cuboid with dimensions a Ò b Ò c the diagonal is:

Glossary.

17 cm

1.2 m

Looking at the red triangle: d = … 2 + 4.8 2 = … m The diagonal of the cuboid measures 5.2 m. Note that you can calculate it directly:

Classify the following triangle as either a rightangled, acute or obtuse triangle. To do this, calculate some of its elements.

Looking at the blue triangle: AC = 1.2 2 + … 2 = … m

5 cm

r = f = 7.07 cm

12 Find the diagonal of a cuboid with dimensions of 1.2 m, 1.6 m and 4.8 m.

Calculate the area of the following squares: 12 m

Calculate the measurements needed to classify the following triangle according to its angles.

10 m

diagonal of 5.8 cm, and one side that measures 4 cm.

c) 2

32 cm

12 cm the perimeter of 7 13 cm Calculate a rectangle with a 20 cm

0 cm

20 cm

15 m 30 cm2

24 cm

10 cm. Remember that the diagonals of the square are perpendicular. Therefore, the coloured triangle is right-angled and has legs of equal length. r 2 + r 2 = …2 8 2r 2 = … 8 r 2 = …

60 m2

14 cm2

10 cm

11 Calculate the radius of the circumscribed circumference of this square with sides of

45 m2

26 cm

r 2 = OP 2 – PT 2 = …2 – …2 = … Acircle = πr 2 = … · … ≈ 643.7 cm2

Calculate the area and the perimeter of these shapes. Note that in the first two the perimeter is the inner and outer periphery.

mm 15

25 m

draw a tangent from P to the circumference. The PT tangent segment is 18 cm. Find the area of the circle. The tangent line is perpendicular to the radius. Therefore, the triangle PTO has a right angle at T :

Find thethe area and perimeter these shapes. Find perimeter of the of following shape:To do so, you will first have to calculate the1unknown length cm of one of their elements. If they are not exact, find them to one decimal place.

12 m

10 The distance of a point P to the centre O of a circumference is OP = 23 cm. We

Exercises and problems. For you to apply all the contents that you learnt throughout the unit.

22 6

20 Calculate the length of x for each of the following Pythagorean theorem

A repeated addition of the same value.

188

Numbers that can be used to count items.

189

187

numeral system

Set of symbols and rules used to represent numbers.

5. DECIMALS

positional numeral system

Numeral system where symbols have different values depending on their level.

subtraction

decimal number

To remove an amount (subtrahend) from another (minuend) to find out the difference between the two.

The result of a non-exact quotient. It has an integer portion and a decimal portion, separated by a decimal point.

hundredth

The result of dividing one tenth into ten equal parts.

number line

A one-dimensional line that contains all the real numbers.

tenth

The result of dividing one into ten equal parts.

thousandth

The result of dividing one hundredth into ten equal parts.

unit

The element used to build all natural numbers, represented by the number 1.

trillón

A 1 followed by 18 zeros (One million billones).

2. POWERS AND ROOTS power

A shortened form of writing a product of equal factors.

power of base 10

The unit followed by as many zeros as figures marked in the exponent.

product of powers with the same base

To multiply two powers with the same base, we keep the same base and add the exponents together.

quotient of powers with the same base

To divide two powers with the same base, we keep the same base and subtract the exponents.

3. DIVISIBILITY

6. THE METRIC DECIMAL SYSTEM angstrom

A unit used to measure atomic distances.

astronomical unit

The average distance from the Earth to the Sun. It is used to measure the distance between planets.

gram

The main unit for measuring masses.

light year

The distance light travels in one year. It is used to measure the distance between galaxies.

divisible

Number that when divided by another gives an exact result.

litre

The main unit for measuring capacities.

composite number

A number that can be factorised into simpler factors.

magnitude

Quality and property of objects that can be measured and quantified numerically.

divisor

Number that is contained in another number an exact number of times.

metric decimal system

The set of units of measurement for basic magnitudes.

factor

Each of the quantities that can be multiplied to form a product.

micrometre

One thousandth of a millimetre.

290

291

MATHS WORKSHOP Unit 9

MATHS WORKSHO READ AND LEARN

PRACTICE MAKES PERFECT!

Pythagoras

Use algebra

Pythagoras (6th century BCE) was known as a mathematician and philosopher. However, his contributions to astronomy are not very well known.

Write any two-digit number and then another number with the same digits swapped round. Subtract one from the other. Can you explain why the difference is always a multiple of 9?

— Moreover, he was one of the first people to realise that some stars, which he called wandering stars, did not have the same regular movement as the other stars. In fact, the word wandering in Greek is pronounced planet. This is why planets were called celestial objects that wander in the sky. We now know that planets are not related to stars. This is why they appear to wander.

INVESTIGATE

• Here is a cross made of four toothpicks.

SELF-ASSESSMENT

b) x

16 m

It is a right-angled triangle. The right angle is at the vertex where the sides with 3 and 4 knots come together.

72 mm

21 mm

shapes: a)

With three stakes, tighten the rope to form a triangle with 3, 4 and 5 knots on the sides.

23 .4

30 m

40 dm

24 cm

30 m

Over 3 000 years ago, the Egyptians used this method to draw right angles.

d) m 31

Every year, after the River Nile’s floods, the borders between the flooded fields needed to be restored.

4 A town square has the shape and the dimensions

shown in the picture. All the angles marked in red are 45°. Calculate the area and perimeter of the square. 12 m

Commitment

4 cm

Remember that you can find academic and professional guidance related to this content at anayaeducacion.es.

40 cm 25 cm

a 40 cm

8.66 m

192

26 cm

26 m

4c m

A bit of history

The land surveyors who were responsible for marking the land borders again used the method shown on the left.

Trust in your skills and knowledge, develop creativity, adapt to changing situations and have a proactive and responsible attitude.

2 Calculate the unknown segment in each of these

Take a rope and mark twelve identical sections by making knots.

Enterprising culture

3 Calculate the area and perimeter of these shapes: 8m

How do you get the right angle in the corners?

Can you make a square just by moving one of the toothpicks?

10 mm

How do we draw the lines? The best way is with a rope pulled tight.

anayaeducacion.es Answer key and interactive self-assessment.

acute-angled or obtuse-angled: a) 20 cm, 24 cm, 30 cm b) 5 m, 6 m, 10 m c) 10 mm, 24 mm, 26 mm d) 7 dm, 7 dm, 7 dm

We want to mark out a beach volleyball court.

Here, you will find readings, activities, advice, information...

Half joking, half serious!

• Here you can see twelve counters arranged in three rows of four. Now arrange them so that there are six rows of four.

1 Classify the following triangles as right-angled,

How to mark out a beach volleyball court

x y – y x

Imagining in space

— He was also the first person to find out that the Moon’s orbit is not in Earth’s equatorial plane, but inclined to it by a certain angle.

Practice makes perfect! In this section you will have to solve many different types of problems.

— He was the first Greek to recognise that the star that we can seen in the morning and at dusk was the same star. We now know that this star is the planet Venus.

x y 8 10x + y y x 8 10y + x

4m 10 m

Watch the video for target 7.2. Think of something you can do to contribute to achieve that goal. Make a commitment to put your idea into practice.

26 m

ment. Self-assess ies, ese activit th By doing r u o y heck you can c how ding and n ta unders t. rn a le e v ha much you

193

Academic and professional

ICT

orientation

Assessment

Linguistic Plan

Learn how to obtain information, select it and apply it; to plan, manage and work on projects; to collaborate online in an ethical and safe manner.

Evaluate your personal skills, discover and awaken your calling, train yourself to make decisions and learn to choose between different options.

Discover different strategies to analyse what you have learnt and how you learnt it; train yourself to take responsibility or overcome difficulties.

Use your communication skills in the different types of text that you will see. Language is always present, communicate!

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SDG

SDG Commitment with short videos that will help you understand the targets for reaching the Sustainable Development Goals worked on in this project. Linguistic Plan with infographics that will give you models to work with the four linguistic skills, using different text types (descriptive, narrative, explanatory, etc.).

Cooperative learning Preparing for the task In small groups, of four or five members: 1 All the members of the team will review how the assigned task can be accomplished. 2 To do this, the steps can be shared out to each team member who then in turn will explain how each part of the process can be done to the others. The others listen and participate if they think they can contribute something.

Authorship / adaptation : Variant of the Educational Innovation Laboratory of the colegio Ártica - David and Roger Johnson.

Developing thinking with explanations are included on how to apply the different thinking techniques proposed in the project.

3 Once everyone is in agreement on how to do each part, you will all complete the tasks and, finally, verify, among everyone, that you have solved it correctly.

Cooperative learning which includes the descriptions of the cooperative learning techniques proposed in the project. Thinking techniques

Emotional education with resources to help you overcome any worries that may arise in different situations at school (beginning of the school year, taking a test, etc.).

Logic Wheel This thinking technique will help you to establish phases when analysing specific content that you have to study.

Identify What is it? What is it like? Are there different types?

By following a logical sequence (the logic wheel), and by asking yourself a series of questions in each phase, you can:

• Identify content by asking yourself: What is it? What is it like? Are there different types? • Compare the content by formulating questions such as: In what way is it similar to ...? In what way is it different from ...? • Establish cause-effect relationships by asking yourself questions such as: Why? What impact does it have ...? • Argue, assess and ask yourself questions such as: What conclusions can be drawn after the analysis? What can be assessed or scored about it? Doing a data dump of these questions into a graphic organiser will help you.

Compare

Argue, assess What can we conclude?

Logic wheel

In what way is it similar to ...? In what way is it different from ...?

3 Establish cause-effect relationships Why? What impact does it have ...?

Authorship: Hernández, P., and García, L. A.; adapted by Escamilla, A.

ICT resources to help you use information and communication technology in a healthy, correct and safe way.

Academic and professional orientation with information on different professions linked to the subject content.

0:40/1:33

Assessment which includes resources for your portfolio, as well as rubrics and targets that will help with your self-assessment.

resources SUBJECT KEY CONCEPTS Tutorials Key concepts Activities with GeoGebra Learn by playing Glossary Self-assessments Language Bank

Resources classified by unit

All the resources are classified by unit so that you can find them more easily.

course contents NATURAL NUMBERS AND INTEGERS

Page 8

1. The set of natural numbers.............................................. 10 2. The relationship of divisibility......................................... 12 3. Prime and composite numbers...................................... 15 4. Lowest common multiple of two or more numbers 16 5. Greatest common divisor of two or more numbers 17 6. The Z set of integers.......................................................... 18 7. Operating with integers.................................................... 19 8. Powers of integers............................................................... 23 9. Roots of integers.................................................................. 25 Exercises and problems......................................................... 26 Maths workshop........................................................................ 30 Self-assessment........................................................................ 31

DECIMAL NUMBERS AND FRACTIONS

Page 32

1. Decimal numbers................................................................. 2. Operating with decimal numbers.................................. 3. Decimal and sexagesimal numbers............................... 4. The square root of a decimal number......................... 5. Fractions.................................................................................. 6. Fractions and decimal numbers..................................... Exercises and problems......................................................... Maths workshop........................................................................ Self-assessment........................................................................

OPERATING WITH FRACTIONS

34 38 42

PROPORTIONALITY

PERCENTAGES

Page 94

1. Percentages. Concept........................................................ 96 2. Problems with percentages............................................. 99 3. Bank interest.......................................................................... 103 4. Other arithmetic problems............................................... 104 Exercises and problems......................................................... 107 Maths workshop........................................................................ 110 Self-assessment........................................................................ 111

ALGEBRA

Page 112

1. Why do we use algebra?................................................... 114 2. Algebraic expressions........................................................ 116 3. Polynomials............................................................................ 119 4. Notable products................................................................. 122 Exercises and problems......................................................... 125 Maths workshop........................................................................ 130 Self-assessment........................................................................ 131

43 44 46 48 54 55

Page 56

1. Adding and subtracting fractions.................................. 58 2. Multiplying and dividing fractions................................. 60 3. Problems with fractions.................................................... 62 4. Powers and fractions.......................................................... 66 Exercises and problems......................................................... 70 Maths workshop........................................................................ 74 Self-assessment........................................................................ 75

Page 76

1. Ratios and proportions...................................................... 78 2. Directly proportional magnitudes................................. 79 3. Inversely proportional magnitudes............................... 82 4. Problems of compound proportionality..................... 84 5. Problems of proportional distribution......................... 86 Exercises and problems......................................................... 88 Maths workshop........................................................................ 92 Self-assessment........................................................................ 92

EQUATIONS

Page 132

1. Equations: meaning and use........................................... 134 2. Equations: elements and terminology......................... 136 3. Transposing terms............................................................... 137 4. Solving simple equations.................................................. 138 5. Equations with denominators......................................... 140 6. The general method for solving first-degree equations................................................................................ 141 7. Solving problems with equations.................................. 142 8. Second-degree equations................................................ 147 9. Solving second-degree equations................................. 148 Exercises and problems......................................................... 150 Maths workshop........................................................................ 156 Self-assessment........................................................................ 157

SYSTEMS OF EQUATIONS

Page 158

1. First-degree equations with two unknowns.............. 160 2. Systems of linear equations............................................. 162 3. Methods for solving linear systems.............................. 163 4. Solving problems with systems of equations........... 166 Exercises and problems......................................................... 171 Maths workshop......................................................................... 176 Self-assessment........................................................................ 177

PYTHAGOREAN THEOREM

Page 178

1. Pythagorean theorem........................................................ 180 2. Calculating a side when two are known..................... 182 3. Applications of the Pythagorean theorem................ 184 Exercises and problems......................................................... 187 Maths workshop........................................................................ 192 Self-assessment........................................................................ 193

SIMILARITY

Page 194

1. Similar shapes....................................................................... 196 2. Plans, maps and models.................................................... 200 3. How to build similar figures............................................. 202 4. Thales’ theorem.................................................................... 204 5. Similarity of right-angled triangles............................... 206 6. Applications of the similarity of triangles.................. 208 Exercises and problems......................................................... 210 Maths workshop........................................................................ 214 Self-assessment........................................................................ 215

GEOMETRIC SHAPES

Page 216

1. Prisms....................................................................................... 218 2. Pyramids.................................................................................. 220 3. Truncated pyramids............................................................ 222 4. Regular polyhedral.............................................................. 224 5. Plane sections of polyhedra............................................ 226 6. Cylinders.................................................................................. 228 7. Cones........................................................................................ 229 8. Truncated cones................................................................... 230 9. Spheres.................................................................................... 233 10. Sections of spheres, cylinders and cones................. 234 Exercises and problems.......................................................... 236 Maths workshop......................................................................... 242 Self-assessment......................................................................... 243

MEASURING VOLUME

Page 244

1. Units of volume..................................................................... 246 2. Cavalieri’s principle............................................................. 248 3. Volume of a prism and a cylinder.................................. 249 4. Volume of a pyramid and a truncated pyramid....... 250 5. Volume of a cone and a truncated cone..................... 252 6. Volume of a sphere............................................................. 253 Exercises and problems......................................................... 255 Maths workshop........................................................................ 260 Self-assessment........................................................................ 261

FUNCTIONS

Page 262

1. The concept of function.................................................... 264 2. Increases, decreases, maximums and minimums.... 265 3. Functions shown in tables of values............................. 266 4. Functions from their equation........................................ 267 5. Proportional functions: y = mx....................................... 268 6. The slope of a line................................................................ 270 7. Linear functions: y = mx + n............................................. 272 8. Constant functions: y = k.................................................. 274 Exercises and problems......................................................... 275 Maths workshop........................................................................ 280 Self-assessment........................................................................ 281

STATISTICS

Page 282

1. Making a table and its graph........................................... 284 2. Location parameters.......................................................... 286 3. Dispersion parameters....................................................... 288 4. Position parameters............................................................ 291 5. Two-way tables..................................................................... 293 Exercises and problems......................................................... 294 Maths workshop........................................................................ 300 Self-assessment........................................................................ 301

CHANCE AND PROBABILITY

Page 302

1. Random events..................................................................... 304 2. Probability of an event....................................................... 306 3. Assigning probabilities to regular experiments....... 308 4. Some strategies for calculating probabilities........... 310 Exercises and problems......................................................... 312 Maths workshop........................................................................ 316 Self-assessment........................................................................ 317

Annex • Glossary.................................................................................. 318

ALGEBRA Reading and listening

The Babylonians, Egyptians, ancient Greeks and Arabs all used rhetorical algebra: they described problems and solutions using everyday language. The Arabs called the unknown the thing. How many sacks of wheat did I gather during the harvest if I sold two in every three, used half of what was left myself and still had five left over?

Some mathematicians like Pythagoras (6th century BCE), Euclid (3rd century CE) and Al Khwarizmi (9th century) used geometry or geometric algebra to demonstrate algebraic relationships and to solve equations. We can use a geometric figure to solve the Egyptian problem above: sold used

Total 8 5 × 2 × 3 = 30

left over

In the 3rd century, Diophantus used a series of abbreviations that simplified algebraic language. He was ahead of his time and no advancements were made in Europe for another twelve centuries. In the 15th century, mathematical terminology developed further, until the French mathematicians Viète (16th century) and Descartes (17th century) finally introduced a wholly symbolic language, virtually indistinguishable from the algebraic language we use today. 1 Create a timeline with the mathematical events described in the page.

Then, write five sentences using for and since.

112

2 Work with a partner. Find five popular abbreviations in English, read

them out loud and have your partner guess what they stand for.

Are abbreviations more popular in writing or when speaking? Discuss. Algebra and geometry 3 What property of addition and multiplication does the following

graphic demonstrate?

m . (a + b + c) = m . a + m . b + m . c

·(

m·a

m·b m

b c

m·c

Algebraic language 4 What is the value of x for the Egyptian problem on the previous page?

• Used 8 1 of x 8 x 3 6 2 x • Left over 8 = 5 6

• N.º of sacks 8 x • Sold 8 2 of x 8 2x 3 3 • Unsold 8 x 3

5 If I build a house of cards with 15 storeys, can you say the number of

holes (triangles) it will contain?

And what about the number of cards? tip: Use the expressions on the right to find the answers. n.° of storeys

n.° of triangles

n.° of cards

…

n 2 n 2 – 2 n2 + n 2

…

3n 2 + n 2

ANK ANK B E G A U LANG LANGUAGE B ANK ANK GE BANK B B E E G G A A U U GUA K LANG ANG N L A L AN GE BANK 113 ANK GE BANK B B E E G G A A U U G A LAN LANG LANGUA LANGU

1 F ocu s on Eng lish branch: an area of knowledge that is considered part of a larger subject of study.

ALGEBRA: WHY DO WE USE IT? Algebra is the branch* of mathematics that uses letters to express unknown, indeterminate or variable numbers. It is a language that makes constructing and describing mathematical processes easier. Let’s see some examples of how algebra is used.

ãã Expressing arithmetic properties • Addition is associative, but subtraction is not. (a + b) + c = a + (b + c) (a – b) – c ≠ a – (b – c) • Multiplication is distributive with respect to addition. a · (b + c) = a · b + a · c

ãã Generalising numerical sequences (general term) anayaeducacion.es Help with reasoning: general term of a sequence.

Example

a1 9

a4 9

a5 … 9

0 2 6 12 20 …

(n – 1) · n

an =

n 2 – n

(0 . 1) (1 . 2) (2 . 3) (3 . 4) (4 . 5) So, if we want to know, for example, the tenth term in the sequence: a10 = 9 · 10 = 90 or a10 = 102 – 10 = 90

ãã Expressing relationships between quantities (formulas) bill Travel costs.............. 8 €50 Cost per hour.......... 8 €35 Time (hours)........... 8 h Materials................. 8 m Total 8 35 · h + m + 50

• The interest, I, that some money in a bank account, C, with an annual rate of r % earns in t months.

8 I = C ·r ·t 12 · 100

Thus, in 8 moths, €4 000 in an account with an 8 I = 4 000 · 1.5 · 8 = €40 12 · 100 annual rate of 1.5 % earns an interest of €40. • The area, A, of a rhombus with diagonals d1 and d2 8 A =

vat 21 %

4 cm

To pay:

6 cm

0.21 · (35 · h + m + 50)

d1 · d2 2

A = 6 · 4 = 12 cm2 2

ãã Expressing and working with numbers of an indeterminate value (algebraic expressions) anayaeducacion.es Consolidate: translating a problem into algebraic language.

114

Examples

• A natural number ÄÄÄÄÄÄÄÄÄÄÄÄ8 • The next number ÄÄÄÄÄÄÄÄÄÄÄÄ8 • Double the next number ÄÄÄÄÄÄÄÄÄ8 • The square of the next number ÄÄÄÄÄÄÄ8

a a+1 2 · (a + 1) (a + 1)2

Unit 6

ãã Expressing relationships that help us to solve problems (Equations) Example

In a 20-question test, 3 marks are awarded for each right answer and 2 are deducted for each question answered incorrectly or left blank. How many questions did Begoña get right and how many did she get wrong if she got a score of 40? Right 8 x Wrong 8 20 – x marks marks total awarded – deducted = score 3x 2 . (20 – x) 40 3x – 2 (20 – x) = 40 8 x = 16 Proof: 3 . 16 – 2(20 – 16) = 40 Solution: Begoña has 16 right answers and 4 wrong answers.

Let’s practise!

1 Which of the identities on the right matches the

definition of the associative property of multiplication: If three or more numbers are grouped in different ways when multiplying, the result will be the same.

a·b·c=c·a·b (a · b) · c = a · (b · c) a · (c + 1) = a · c + a

2 Copy and complete the empty boxes. 1

4 10

. . .

3n – 2

3 Write the first five terms of a sequence with the general

term an = 3n + 1 . 2 4 Write the general term of these sequences:

6 The sum of the first n natural numbers is: 2 1+2+3+4+…+n= n +n 2 Calculate the sum 1 + 2 + 3 + … + 50.

7 A clothes shop owner buys 100 t-shirts in order to sell

them. To plan his accounts, he manages the following variables: C 8 Total cost of the t-shirts s 8 Selling price (per unit) E 8 Expenses P 8 Profit Write an equality that relates these four variables.

8 Translate the ages of these family members into

algebraic language in your notebook:

a) 1 - 4 - 9 - 16 - 25 - … 8 an = ? b) 0 - 3 - 8 - 15 - 24 - … 8 bn = ? 5 The gross monthly salary, income tax (IRPF) and net

salary of a company’s employees are calculated using the following formulas: Sb = 900 + 3a + 10b Tax = 0.21 · Sb Sn = 0.79 · Sb

a = Years worked b = Overtime

a) Calulate the salary of an employee who has worked for 8 years and done 21 hours of overtime. b) How much income tax will he pay?

age

Javi He is x years old.

Pepa (sister) She is one year younger than Javi. Carol (mother) Javi was born when she was 22. Álex (father) He is triple the age of Pepa. 9 Write an equality about the family in the previous

exercise that reflects this new data: Javi’s father is 3 years older than his mother. Calculate Javi’s age by trial and error.

anayaeducacion.es Write the elements of a sequence. 115

ALGEBRAIC EXPRESSIONS The type of expressions that are formed with letters and numbers are called algebraic expressions. Let’s start by looking at the simplest type: monomials.

ãã Monomials A monomial is the indicated product of a known value (coefficient) and one or various unknown values, represented by letters (literal part).

Example –5ax 2

coefficient

⎯8 monomial coefficient Ä8 –5 literal part Ä8 ax 2

3 anayaeducacion.es Practise recognising the elements of monomials.

literal part

coefficient

literal part

➜➜ degree of a monomial

The number of factors that form the literal part is called the degree of the monomial.

degree Ä8 3 a·x·x

3 xy 2 5

4a 2 8

second-degree monomial

5x 2y 2 8

a · a

fourth-degree monomial

x·x·y·y

➜➜ numerical value of a monomial

This is the value of the monomial when the letters take specific values. The numerical value of 2ab 2 when a = 1 and b = 2 is 8.

a=1

2ab 2 ⎯⎯8 2 · 1 · 22 = 8 b=2

➜➜ similar monomials Example

We say that two monomials are similar when they have the same literal part.

5x 2 ←⎯8 1 x 2 2 identical literal part

anayaeducacion.es Practise adding and subtracting monomials.

are 3a ⎯⎯⎯8 –2a similar

are 1 x 2y 4x 2y ⎯⎯⎯8 similar 5

ãã Adding monomials • Two monomials can only be added if they are similar. In this case, the

coefficients are added together, leaving the literal part the same. • If the monomials are not similar, the sum does not change. Examples

• 5a + 2a = 7a • 8x 2 – 3x 2 = 5x 2 • 3x + 2x 2 ÄÄ8 no change • a 2 – a + a 2 = 2a 2 – a ÄÄ8 no change 116

Unit 6

eas

id Consolidating

1 Copy and complete in your notebook.

a) 2x + 3x = 5x b) 5x – 2x = 3x c) 3x 2 + 4x 2 = 7x 2 d) 7x 2 – 4x 2 = 3x 2 3x + x = … 3x – x = … 6x 2 + x 2 = … 5x 2 – 2x 2 = … 5x + 4x = … 7x – 3x = … 2x 2 + 6x 2 = … 5x 2 – 4x 2 = … 3x + 4x + x = … 2x – 3x = … 5x 2 + x 2 + 2x 2 = … 3x 2 – 7x 2 = … 2 Copy in your notebook and simplify.

a) 4x + x – 6 + 2 = 5x – 4 b) 4x 2 – 2x 2 + 7 + 1 = 2x 2 + 8 c) x 2 + 2x 2 + 7x – 2x + 1 = 3x 2 + 5x + 1 x + x – 3 + 5 = 2x + … 6x 2 – 5x 2 + 3 + 4 = x 2 + … x 2 + 3x 2 + x – 5x + 6 = 4x 2 – … + 6 3x – 2x + 2 + 2 = … + 4 5x 2 – 3x 2 – 2 – 2 = … – 4 9x 2 – 2x + 3x – 3 + 1 = … + x – … 8x – 5x – 3 – 2 = … 6x 2 + 2x 2 + 3 – 6 = … 7x 2 + 3x 2 + 7x – 2 – 4 = … + … – … Let’s practise! 1 Copy and complete in your notebook. monomial

–3x

a 2b

7 Use the example as a model and simplify. 2 xy 4 3 1 4 ab

coefficient literal part degree

2 Add the following monomials:

a) a + a b) m + m + m c) x + x + x d) n+n+n+n e) x 2 + x 2 f ) a 3 + a 3 + a 3 + a 3 3 Add the following expressions:

a) 4a + 2a c) 3x 2 + 6x 2 e) m 3 + 2m 3 + 4m 3

b ) 4m + 4m d) 5a 2 + a 2 + 2a 2 f ) 3x 4 + 6x 4 + 2x 4

8 Problem solved

Remove the brackets and simplify. a) (5x + 1) – (2x – 3) = 5x + 1 – 2x + 3 = 3x + 4 b) (4x 2 – 6) – (x 2 – 2x + 1) = 4x 2 – 6 – x 2 + 2x – 1 = = 3x 2 + 2x – 7 9 Remove the brackets and simplify.

4 Problem solved

Add the following expressions: n 3 + 2 n3 = 9 n3 a) 1 x + x = 4 x b) 10 2 3 3 5 5 Simplify by adding.

n 2 + 2n 2 3m + 2m c) a) x + 1 x b) 4 3 7 7 2

6 Subtract these monomials:

a) 8x – 3x b) 8a – 7a 2 2 d) 5a – 9a e) m 3 – 5m 3 3a 2 – 1 a 2 g) 5 x – 1 x h) 6 2 4 6

• 3x + 6 + x + 2 = 3x + x + 6 + 2 = 4x + 8 a) 3a + 3 – 2a + 1 b) 5x + 2 – 3x + x c) 7 – 4a – 7 + 5a d) 4x – 3 – 4x + 2 2 2 e) x + 4 + x + 1 f ) 5x 2 – 3 – 4x 2 + 1 g) x 2 + 4x + 1 + 2x + 3 h) 5x 2 + 3x – 4x 2 – 2x + 1 i) 3x 2 + 4 – x 2 + 2x – 1 5 5 j) 10 – 3 x + 1 x 2 – 7 – x 2 2

c) 11x 2 – 6x 2 f ) 4n 4 – n 4 3 3 i) a – 2a 2 5

a) 3x + (2x – 1) b) 7x – (5x – 4) c) 6x – (4x + 2) d ) 3x – (x + 5) e) (x – 5) + (x – 3) f ) (4x + 2) – (3x + 2) 2 2 g) (3x – 5x + 2) + (x – 2x + 1) h) (5x 2 – 2x – 3) – (4x 2 + 3x – 1) i) (x – 3) + (x 2 + 2x + 1) j) (6x 2 – x) – (3x 2 – 5x + 6) 10 Calculate.

a) Numerical value of 5x 2 when x = 1. b) Numerical value of – 4x 2 when x = –3. c) Numerical value of –2xy when x = 3 and y = –5.

anayaeducacion.es GeoGebra. Adding polynomials. 117

2 ALGEBRAIC EXPRESSIONS

ãã Multiplying monomials As you know, a monomial is a product of numbers and letters. Therefore, the product of two monomials is another monomial.

Observe

Examples

third-degree second-degree

• (3a) · (2a) = 3 · 2 · a · a = 6a 2 • (5x) · (–3x 2) = 5 · (–3) · x · x 2 = –15x 3

(2x 2) · (3x 3) = 6x 5 fifth-degree

• (3a) · d 5 abn = 3 · 5 · a · a · b = 15 a 2b = 5 a 2b 6 6 6 2

The degree of the product is equal to the sum of the degrees of the factors.

ãã Dividing monomials The coefficient of two monomials can be a number, another monomial or an algebraic fraction. Examples

Interesting fact

2 • 10a 2: 5a 2 = 10 · a = 2 ÄÄ8 (number) 5a 2

An algebraic fraction is a fraction with a literal part in the denominator. 5a 8 Algebraic fraction 3b 2 4x = 1 8 Algebraic fraction 12x 2 3x

2 • 18x 3 : 3x 2 = 6 $ 3 · x · x = 6x ÄÄ8 (monomial) 3x 2

• (2a2) : (6ab) =

2 · a · a = a ÄÄ8 (fraction) 2 · 3 · a · b 3b

Given that the letters represent numbers, operations with algebraic expressions keep all the properties of numerical operations.

anayaeducacion.es Practise multiplying and dividing monomials.

eas

id Consolidating

3 Copy and complete in your notebook.

a) (ab) · (ab2) = a

d) x : x 2 = x2 = 1 d x

b) (–2ab) · (ab2) = – a b

c) (–2ab) · d 3 ab 2n = – d a b 4 2

e) 2x : 6x 2 = 2x2 = d 3x 6x

f ) 2xy 2 : 6x 2 y =

y 2xy 2 = 2 6x y 3d

Let’s practise! 11 Do the following multiplications:

a) (3x) · (5x)

b) (–a) · (4a)

2 d x n · (6x) c) (4a) · (–5a 2) d) 2

e) e x o · d x n 3 2 2

f ) (5a) · d– 1 a 2n 5

12 Multiply these monomials:

a) (3x) · (5xy)

b) (–2ab) · (4b)

d– 2 abn · d– 3 abn c) (4x 3y) · (xy) d) 3 2

13 Simplify.

5x 3 c) a) 4x b) 2 3a 10x 2 15x d) 12a e) 4a 3x 2

2 f ) 8a 3 8a

14 Divide.

a) (10x) : (2x) c) (14a 2) : (–7a) e) (10x 2) : (5x 3) g) (–16a 4) : (8a 6)

b) (5a 2) : (15a 2) d) (6x 3) : (9x 2) f ) (–5a) : (–5a 3) h) (27x 3) : (–9x)

anayaeducacion.es GeoGebra. Simplifying algebraic fractions with monomials. 118

Unit 6

POLYNOMIALS • The sum (or subtraction) of two monomials is a binomial. • The sum (or subtraction) of three monomials is a trinomials. • In general, the sum (or subtraction) of multiple monomials is a polynomial. Examples

x+y

a2 – 1

x 2 – 3x + 1

binomials

a 2 – ab + 2

5x 4 – 3x 3 + 2x – 1

trinomials polynomials

➜➜ degree of a polynomial

The degree of a polynomial is the largest of the degrees of the monomials that form it.

Observe polinomial degree 8x 3 – 5x 2 + 7

8 3

5 – 4x 3 x

8 5

+ 1

3x 2 + 11x

Example

2x 4 – 5x 2 + 3x – 8 Ä8 fourth-degree polynomial 4th degree 2nd degree 1st degree 0 degree

8 2

➜➜ numerical value of a polynomial

When the letters take specific values, the polynomial also takes a specific value. Example

For the polynomial 3x 2 – 2x + 5: • When x = 0 8 3 · 02 – 2 · 0 + 5 = 0 – 0 + 5 = 5 The numerical value of 3x 2 – 2x + 5 when x = 0 is 5. • When x = –2 8 3 · (–2)2 – 2 · (–2) + 5 = 12 + 4 + 5 = 21 The numerical value of 3x 2 – 2x + 5 when x = –2 is 21. The numerical value of a polynomial depends on the value of the letters. ➜➜ opposite of a polynomial

The opposite of a polynomial is another polynomial which has the same monomials, but with the signs changed: plus to minus and minus to plus. Example

opposite polynomials

7x 3 + 3x 2 – 5x + 8

5 –7x 3 – 3x 2 + 5x – 8

Let’s practise! 1 Indicate the degree of each polynomial.

a) x 2 – 3x + 7 b) x 4 – 2 d) 9x 6 + 2x e) x 5 – 2x 2

c) 5x 3 – 3x 2 f ) 6x 4 – 3x 4

2 Calculate the numerical value of x 3 – 5x 2 – 11:

a) When x = 1.

b) When x = –1.

3 Calculate the values of x that cancel out each

polynomial through trial and error. a) x 2 – 2x + 1 b) x 3 – 8

c) x 4 – x 3

4 Give the opposite in each case:

a) x 3 – 5x + 1

b) 2x 4 + 6x 3 – 8x 2 + 3x – 1

119

3 POLYNOMIALS

ãã Adding polynomials To add two or more polynomials, we must remember what we already know about adding monomials.

General rule To add two (or more) polynomials, one is placed under the other with the similar monomials in the same column.

For example, let’s add the polynomials A = 2x 3 – 3x 2 + 6 and B = x 2 – 5x + 4. • With what we already know, we can do the following: A + B = (2x 3 – 3x 2 + 6) + (x 2 – 5x + 4) = 2x 3 – 3x 2 + 6 + x 2 – 5x + 4 = = 2x 3 – 3x 2 + x 2 – 5x + 6 + 4 = 2x 3 – 2x 2 – 5x + 10 • In practice, it is usually done like this: A 8 2x 3 – 3x 2 + 0x + 6 x 2 – 5x + 4 B 8+ A + B 8 2x 3 – 2x 2 – 5x + 10

anayaeducacion.es Practise adding and subtracting polynomials.

ãã Subtraction of polynomials Let’s subtract the polynomials A and B we saw above. General rule

• With what we already know, we can do the following:

To subtract two polynomials, add the first to the opposite of the second. In other words, change the sign of the second and add them together.

A – B = (2x 3 – 3x 2 + 6) – (x 2 – 5x + 4) = 2x 3 – 3x 2 + 6 – x 2 + 5x – 4 = = 2x 3 – 3x 2 – x 2 + 5x + 6 – 4 = 2x 3 – 4x 2 + 5x + 2 • In practice, it is usually done like this: A 8 2x 3 – 3x 2 + 0x + 6 –x 2 + 5x – 4 –B 8 + A – B 8 2x 3 – 4x 2 + 5x + 2

eas

id Consolidating

1 Copy and complete in your notebook.

x 2 + 5x – 7 b) 3x 3 – 6x 2 + 8x + 2 c) – x 2 + 3x – 9 d) x 3 – 4x 2 – – 1 + x 2 – 8x + 5 + 2x 3 + 2x 2 – 6x – 9 + – x + + – + x+ 2 2 3 2 – –2 – 4x + – 3x + 2x – 5 3x – 6x – 5x + 3

2 Given the polynomials P = x 4 + 5x 3 – 7x – 6 and H = x 3 – 4x 2 – x + 8, copy and complete.

P 8 x 4 + 5x 3 + 0x 2 – 7x – 6 x 3 – 4x 2 – x + 8 H 8+ P + H 8 x 4 + – – +2

P 8 x 4 + 5x 3 + 0x 2 – 7x – 6 –x 3 + 4x 2 + x – 8 –H 8 + P – H 8 x 4 + + – – 14

Let’s think: Why have we added the addend 0x 2 to polynomial P? Let’s practise! 5 Calculate

the following operations with these polynomials: A = 3x 3 – 5x 2 – 4x + 4 B = 2x 3 – x 2 – 7x – 1 a) A + B b) A–B

6 Calculate

the following operations with these polynomials: M = 7x 3 – 6x 2 + 2 N = 5x 2 – 3x – 5 a) M + N b) M – N c) N–M anayaeducacion.es Adding and subtracting polynomials.

120

Unit 6

ãã Product of a polynomial and a monomial To multiply a polynomial by a monomial, we multiply the monomial by each of the addends of the polynomial (distributive property). We first multiply the polynomial x 3– 4x 2 + 5x – 1 by a number (2), then by a first-degree monomial (–3x) and finally by a second-degree monomial (x 2). • (x 3 – 4x 2 + 5x – 1) · 2

• (x 3 – 4x 2 + 5x – 1) · (–3x)

x 3 – 4x 2 + 5x – 1 2 × 3 2 2x – 8x + 10x – 2

x 3 – 4x 2 +

5x – 1 – 3x × 4 3 2 –3x + 12x – 15x + 3x

• (x 3 – 4x 2 + 5x – 1) · x 2 x 3 – 4x 2 + 5x – 1 x 2 × x 5 – 4x 4 + 5x 3 – x 2

ãã Product of polynomials General rule

By combining the products above, we can calculate the product of two polynomials using a similar procedure to the one we use to multiply numbers.

To calculate the product of two polynomials, we multiply each monomial of one factor by each of the monomials of the other factor and add all the monomials obtained, simplifying those that are similar.

x 3 – 4x 2 x 2 × 2x 3 – 8x 2 – 3x 4 + 12x 3 – 15x 2 + x 5 – 4x 4 + 5x 3 – x 2 x 5 – 7x 4 + 19x 3 – 24x 2

anayaeducacion.es Practise multiplying polynomials.

+ – + +

5x – 1 Z 3 ] (x – 4x 2 + 5x – 1) · (x 2 – 3x + 2) = 3x + 2 ] = (x 3 – 4x 2 + 5x – 1) · x 2 + 10x – 2 8 [ 3 2 3x ] + (x – 4x + 5x – 1) · (–3x) + ] + (x 3 – 4x 2 + 5x – 1) · 2 \ + 13x – 2

(x 3 – 4x 2 + 5x – 1) · (x 2 – 3x + 2) = x 5 – 7x 4 + 19x 3 – 24x 2 + 13x – 2

eas

id Consolidating

3 Copy and complete the following multiplications:

a) (2x 3 – 3x 2 + 5x + 6) · (x – 2) b) (4x 3 + 5x 2 + 3) · (x 3 – 3x +1) 2x 3 – 3x 2 + 5x + 6 4x 3 + 5x 2 x– 2 × × x 3 + 0x 2 – 4x 3 + – – 12 4x 3 + 5x 2 – 3x 3 + + 6x – – – + 6 3 + + 4x + + 3x – + – – 12 + – – +

+ – + –

0x + 3 3x + 1 0x + 3 9x

–

+ 3

Let’s practise! 7 Calculate.

a) 3 · (2x + 5) c) 7 · (x 3 – 1) e) x · (x + 1) g) x 2 · (5x – 2) i ) 3x · (x 2 – 2) k) (–2x) · (x 2 + 3)

8 Multiply.

(x 2

b) 5 · – x) d) (–2) · (5x – 3) f ) 2x · (3x – 5) h) 3x 2 · (x + 2) j ) 5x · (x 2 + x + 1) l ) –x · (x 3 + x + 3)

a) (x + 1) · (x – 2) c) (2x – 3) · (3x – 2)

b) (2x – 1) · (x – 1) d) (4 + x) · (2x + 1)

9 Calculate the following products:

a) (2x + 1) · (x 2 – x – 1) b) (3x 2 – 2) · (2x 2 + 4x – 3) c) (x 3 + 2x 2 – 3) · (3x 2 + 5x – 4)

anayaeducacion.es Product of polynomials. 121

NOTABLE PRODUCTS Notable products are certain products of binomials that are useful to remember because they can help us to do calculations with algebraic expressions more quickly.

anayaeducacion.es Practise calculating notable products.

ãã Square of a sum a+b

Have a look at the following: b

a+b

a·b

• (x + 1)2

x × x x + x 2 + x x 2 + 2x

+ 1 + 1 + 1 + 1

• (x + 5)2

x × x 5x + 5x 2 + 5x 5x 2 + 10x

+ 5 + 5 + 25

• (3x + 2)2

+ 25

3x × 3x 6x + 9x 2 + 6x 9x 2 + 12x

+ 2 + 2 + 4 + 4

As a general rule: b2

a·b

3x + 4 3x + 4 12x + 16

9x 2

+ 12x

9x 2 + 24x + 16

a+ × a+ ab + 2 + a + ab a 2 + 2ab +

The square of a sum of two monomials is equal to:

b b b 2

the square of the first addend, plus double the first multiplied by the second, plus the square of the second. (a + b)2 = a 2 + 2ab + b 2

b 2

The statement above is true for any addends a and b, so it can be applied automatically without having to do the multiplication. Example

(3x + 4)2 = (3x)2 + 2 · 3x · 4 + 42 = 9x 2 + 24x + 16

ãã Square of a difference Like with the square of a sum:

3x – 4 3x – 4

– 12x + 16 2 + 9x – 12x 9x 2 – 24x + 16

a– × a– – ab + 2 + a – ab a 2 – 2ab +

The square of a difference of two monomials is equal to:

b b b 2

the square of the first addend, minus double the first multiplied by the second, plus the square of the second.

b 2

(a – b)2 = a 2 – 2ab + b 2

Example

(3x – 4)2 = (3x)2 – 2 · 3x · 4 + 42 = 9x 2 – 24x + 16

ãã Difference of squares ×

3x + 4 3x – 4

– 12x – 16 2 + 9x + 12x 9x 2 + 0 – 16

× – 2 + a + a 2 +

a a ab ab 0

+b –b – b 2 – b 2

The sum of two monomials multiplied by their difference is equal to the difference of their squares. (a + b) · (a – b) = a 2 – b 2

Example

(3x + 4) · (3x – 4) = (3x)2 – 42 = 9x 2 – 16 anayaeducacion.es Expand notable products.

122

Unit 6

ãã Applications of notable products Notable products can be used, for example, when factorising some polynomials and when simplifying fractions. Examples

anayaeducacion.es Identifying notable products.

• Factorising the polynomial x 2 – 4x + 4: x 2 – 4x + 4 = x 2 – 2 · 2 · x + 22 = (x – 2)2

double the first multiplied by the second

square of the first

square of the second

• Factorise x 2 – 4: x 2 – 4 = x 2 – 22 = (x + 2) · (x – 2)

Remember ab = a $ b b2 b $ b

difference of = sum · difference the squares

= a b

• Keeping this in mind, we can simplify the following fraction:

a = a$1 = 1 ab b a$b

x 2 – 4 = (x + 2) · (x – 2) = (x + 2) · (x – 2) = x + 2 ( x – 2) · ( x – 2) x – 2 x 2 – 4x + 4 ( x – 2) 2

eas

id Consolidating

1 Copy and complete in your notebook. Multiply using the formulas from the previous page

and then check by doing the operation. a) (2x + 7)2 = (2x)2 + 2 · 2x · 7 + 72 = b) (2x – 7)2 = (2x)2 – 2 · 2x · 7 + 72 = = + + = – +

2x + 7 × 2x + 7 + + + +

c) (2x + 7) · (2x – 7) = (2x)2 – 72 = = –

2x – 7 × 2x – 7 – + – – +

2x + 7 × 2x – 7 – – + + –

Let’s practise! 1 Copy and complete.

3 Copy and complete.

a) (x + 1)2 = x 2 + 2 ·

b) (x – 5)2 = x 2 – 2 ·

+ 52 = x 2 –

c) (x + 5) · (x – 5) =

2 = x 2 + 2

2 – 52 = x 2 –

Check the results by calculating each product. 2 Calculate.

a) (x +

4)2

d) (a + 2)2 g) (2x –

y)2

b) (x –

1)2

e) (a – 1)2 h) (5 –

3x)2

c) (x – 6) · (x + 6) f ) (a + 4) · (a + 4) i) (2x + 1) · (2x – 1)

a) a 2 – 1 = (a + 1) · ( – ) b) a 2 – 2a + 1 = ( – )2 c) a 2 – 16 = ( + ) · ( – d) x 2 + 2xy + y 2 = ( + )2

)

4 Simplify the following fractions:

a) c)

x 2 + 2xy + y 2 a2 – 9 b) 2 2 2 x –y a – 6a + 9 a2

a 2 – 1 d) a 2 – 16 a+4 – 2a + 1

anayaeducacion.es Practise simplifying fractions. 123

4 NOTABLE PRODUCTS

ãã Taking out the common factor Taking out the common factor refers to a transformation that can be applied to certain additions and subtractions, and is very useful in algebraic calculations. Look at the following expression: anayaeducacion.es Practise taking out the common factor.

a·b+a·c–a·d

— It is a sum whose addends are products. — All the products have the common factor a.

Therefore, we can transform the sum into a product by taking out the common factor and placing the rest in brackets. a · b + a · c – a · d = a · (b + c – d )

A special case If the common factor to be removed is the same as one of the addends, this addend is replaced by the number one. a + ab = a · 1 + ab = a · (1 + b)

Note that this transformation is just the application of the distributive property. Examples

• a 2 + ab = a · a + a · b = a · (a + b)

• 4 · a + 4 · b = 4 · (a + b)

• x 3 – 2x 2 + 5x = x 2 · x – 2x · x + 5 · x = (x 2 – 2x + 5) · x ➜➜ applications

We take out the common factor to simplify fractions, as you can see in the following examples: Examples

Remember

• 5a2 + 5b = 5 · (a + b) = 5 a + ab a · (a + b) a

Explain the steps followed to simplify: 5 = 5 ·1 = 1 5a + 5 5 · (a + 1) a + 1

•

x3 = x2 · x = x x 2 + x 3 x 2 · (1 + x) 1 + x

2 = m · (m – n) = m • m 2 – mn 2 (m + n) · (m – n) m + n m –n

eas

id Consolidating

2 Copy and complete.

3 Copy and complete like in the example.

a) 7x + 7y = 7 · (x + )

• 2a · (a + 3) = 2a 2 + 6a ↔ 2a 2 + 6a = 2a · (a + 3)

b) ax – ax = a · ( – y)

a) 5a · (2 + a) =

c) 6a – 9b =

b) 3x · (1 – 4x) =

· (2a – 3b)

d) x 2y – xy 2 = xy · ( – )

c) x 2 · (x – 5) =

+ – –

↔ 10a + 5a 2 = 5a · ( + ) ↔ 3x – 12x 2 = ↔ x 3 – 5x 2 =

·( – ) ·( – )

4 Have a look at the Remember box. Then, copy and complete in your notebook. a)

x x $1 = =… 2 x ( d + d) x + 3x

3a = 3a =… 2 a + 1) d ( a +a

2x 2 = 2 $ x2 =… 6x 2 + 2x 2x (d + d)

Let’s practise! 5 Take out the common factor.

a) 8x + 8y d ) 2a 2 + 6a

6 Simplify.

c) x 2

b) 8 + 4a + xy 3 3 e ) 6a + 2a f ) x + x 2 – x

3x b) x2 4a c) 4a + 8b 2x + xy x 2 + x3

anayaeducacion.es Simplifying algebraic fractions with polynomials. 124

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from this un

Unit 6

MS ES AND PROBLE

EXERCIS

Using algebraic language 1

Write an algebraic expression for the following statements, taking x as any number.

Copy and complete in your notebook. 1

b) Half of the previous number.

f ) A number 5 units greater than triple x. 2

On a farm there are H horses, C cows and D ducks. Match these expressions with the number of: a) Legs. b) Heads. c) Ears. A 2H + 2C

B H+C+D

C 4(H + C) + 2D

If x is the monthly salary of an employee, express algebraically:

b) If you triple Jorge’s age, x, and add 5 years to the result, you get his father’s, who was 33 years old when Jorge was born. Jorge’s age 8 x 5

a) … a three-digit number a b c . b) … the next number? c) … its double? d) … double the number before it? A 100a + 10b + (c + 1) B 200a + 20b + 2c C 200a + 20b + 2c – 2 D 100a + 10b + c

…

n n (n + 1) 2

…

–5

Following the logic of the table, complete the empty boxes in your notebook. 1

399 10

Write the expression for the nth term in each of these sequences:

d) 4 - 9 - 14 - 19 - 24 - … 8 dn = ? 9

The nth term of a sequence is given by the expression: an = 5n – 4 a) Write down the first five terms. b) What is the value of a100?

The nth term of a sequence is given by the expression: an = 3n – 1 2 Calculate the terms a5, a9 and a15.

Copy and complete the table in your notebook, where the values a, b and c are related to each other by the formula: a = 3b + 2c 5

His father’s age 8 x + 33

Which of the following expressions represents…

…

c) 5 - 10 - 15 - 20 - 25 - … 8 cn = ?

Write each of these statements as an algebraic equality: a) If you increase a number, x, by 15 units and divide the result by 2, you get triple that number.

n 3n 2

b) 3 - 5 - 7 - 9 - 11 - … 8 bn = ?

b) The total wage in December, when employees are given the bonus payment.

…

a) 2 - 4 - 6 - 8 - 10 - … 8 an = ?

a) The value of a bonus payment, assuming it is equivalent to 80 % of the monthly salary.

c) The annual income, if there are two bonus payments: in summer and at Christmas.

4 10

c) The result of adding 3 units to x. e) 3 times the number that results from adding 5 units to x.

a) Triple x.

d) Half of a number 3 units greater than x.

b c a

0 0

0 5

2 7

3 3

4 9

125

EXERCISES AND PROBLEMS 12

Each of these tables follows the same logic. The relationship between the numbers in each box is the same. Complete them in your notebook. 2A – B

A2 – B2

A·B

7 3

21 13

10 1

b) 3x – x 2 + 5x + 2x 2 – x – 1

2 5

10 12

– 6

d) 5x 3 – 1 – x + x 3 – 6x 2 – x 2 + 4

d) 7x 4 – x 3 + x 2 + 1

Simplify.

c) 2x 2 + 4 + x 3 – 6x + 2x 2 – 4 Remove the brackets and simplify. a) (3x 2 – 5x + 6) + (2x – 8)

b) (6 – 3x + 5x 2) – (x 2 – x + 3)

2 xy 3

c) (9x 2 – 5x + 2) – (7x 2 – 3x – 7)

coefficient

literal part

a 3b

d) (3x 2 – 1) – (5x + 2) + (x 2 – 3x) 20

a) 2x + 8x

b) 7a – 5a

c) 2x – 5x

d) 3a – 10a

e) 8x – 6 – 3x – 1

f ) 6a – 2 – 5a – 1

g) 2x + 3 – 9x + 1

h) a – 6 – 2a + 7

b) 3x + (2x + 3)

c) (5x – 1) – (2x + 1)

d) (7x – 4) + (1 – 6x)

e) (1 – 3x) – (1 – 5x)

f ) 2x – (x – 3) – (2x – 1) b) 12x : 3x

c) x 2 · x 3

d) 15x 6 : 5x 4

f ) (–20x 8) : 5x 7 3x 2 : x g ) (–2x 2) · (3x 4) h) 4 4 i) 2x · 6x j ) x 2 : x 5 3 3 2 k) 3x · (–3x 3) l ) 2x : (–2x 3) 5 4 2 2x : x 3 m) x · 2x n) 2 3 3 6

Look at these polynomials and calculate: B = x 3 – 3x + 1

C = 2x 2 + 4x – 5 a) A + B b) A + B + C c) A–B d) B – C e) A + B – C 22

Calculate. a) 2 · (x 3 – 3x 2 + 2x + 2)

Calculate and simplify. a) 3x · 4x

x 3 – 3x 2 + x – 8 + 4x 3 + x 2 – 5x – 6x 3 + 2x 2 – x – 10

A = 3x 3 – 6x 2 + 4x – 2

Remove the brackets and simplify. a) x – (x – 2)

Copy and complete. 3x 2 – 5x – 5 + x 2 + x – 5x 2 – x – 6

Calculate.

b) (– 4) · (2x 2 – 5x – 1) c) x · (3x 3 – 4x 2 – 6x – 1) d) x 2 · (5x 2 + 3x + 4)

e) 3x · 5x 3

126

c) 2x 5 – 4x 2 + 1

a) x 2 – 6x + 1 + x 2 + 3x – 5

degree

b) 4 – 3x 2

monomial

a) x 3 + 3x 2 + 2x – 6 18

Copy and complete.

Write the degree of each of the following polynomials:

A B

Monomials 13

Polynomials

e) (–2x) · (x 3 – 2x 2 + 3x + 2) 23

Simplify. a) 2(3x – 1) + 3(x + 2) b) 3(x 2 – 2x – 1) – 2(x + 5) c) 4(2x 2 – 5x + 3) – 3(x 2 + x + 1) d) 6(3x 2 – 4x + 4) – 5(3x 2 – 2x + 3)

f ) A – B – C

Unit 6

Multiply. a) (x – 1) · (2x – 3) b) (3x – 2) · (x – 5) c) (2x + 3) · (3x – 4) d) (x + 1) · (x 2 + x + 1) e) (3x + 2) · (x 3 – 2x 2 + 5x + 1) f ) (x 2 – 2x – 3) · (2x 3 – 5x 2 – 4x + 3)

Notable products and taking out the common factor 31

Problem solved

Multiply: (x 3 – 5x + 1) · (x 2 + 3) x 3 + 0x 2 – 5x + x 2 + 0x + × 3 3x + 0x 2 – 15x + 5 4 + x + 0x – 5x 3 + x 2 x 5 + 0x 4 – 2x 3 + x 2 – 15x +

1 3 3

When the polynomials being multiplied are incomplete, we include the monomials whose coefficients are zero. 26

Calculate. a) (x 2 + 1) · (x – 2) b) (2x 2 – 1) · (x 2 + 3) c) (2x – 3) · (3x 3 – 2x + 2) d) (x 2 + 2) · (x 3 – 3x + 1)

Calculate like in the example. • (x 2 + 3) · (x 2 – 1) = x2 · (x 2 – 1) + 3 · (x 2 – 1) = = x 4 – x 2 + 3x 2 – 3 = x 4 + 2x 2 – 3 a) (x + 1) · (x 2 + 4) b) (x 3 + 1) · (x 2 + 5) c) (x 2 – 2) · (x + 7) d) (x 3 – 3x + 5) · (2x – 1) Simplify. a) (x + 1) · (2x + 3) – 2 · (x 2 + 1) b) (2x – 5) · (x + 2) + 3x · (x + 2) c) (x 2 – 3) · (x + 1) – (x 2 + 5) · (x – 2) d) (4x + 3) · (2x – 5) – (6x 2 – 10x – 12)

Problem solved

Divide: (9x 3 – 15x 2 + 6x) : 3x 2 3 (9x 3 – 15x 2 + 6x) : 3x = 9x – 15x + 6x = 3x 3x 3x 2 = 3x – 5x + 2

Do the following divisions: a) (8x – 6) : 2 b) (20x – 5) : 5 2 c) (3x – x) : x d) (4x 3 – 8x 2) : 2x e) (4x 3 – 2x 2 + 6x) : 2x f ) (12x 3 + 9x 2) : 3x 2

Take out the common factor. a) 3x + 3y + 3z

b) 2x – 5xy + 3xz

c) a 2

d) 3a – 6b

+ 3a

e) 2x + 4y + 6z

f ) 4x – 8x 2 + 12x 3

g) 9a + 6a 2 + 3a 3

h) 2a 2 – 5a 3 + a 4

Calculate using the formulas for the notable products, without multiplying. a) (x + 3)2

b) (3 + a)2

c) (2 – x)2

d) (a – 6)2

e) (2x + 1)2

f ) (5 – 3a)2

g) (x – 5) · (x + 5)

h) (3x – 5) · (3x + 5)

Problem solved

Factorise these expressions: a) x 2 – 8x + 16 = (x – 4)2 = (x – 4) · (x – 4) b) x 3 – 4x = x · (x 2 – 4) = x · (x + 2) · (x – 2) c) 5x 2 + 10x + 5 = 5 · (x 2 + 2x + 1) = 5 · (x + 1)2 = 34

= 5 · (x + 1) · (x + 1) Factorise.

a) x 2 – 6x + 9

b) x 3 – 9x

c) 3x 2 + 6x + 3

d) 2x 3 – 12x 2 + 18x

e) x 4 – x 2

f ) 4x 2 + 4x + 1

Take out the common factor in the numerator and the denominator, then simplify.

2x 2 + 10x c) 2x 2 – 2x x b) 3x 3 + 15x 2 2x 3 x 2 + 2x

Factorise the numerator and denominator, then simplify.

5x + 15 x 2 – 9 b) – 6x + 9 x 2 + 6x + 9

x 2 + 2x + 1 c) 3x2+ 3 d) 5x 2 + 5x 3x – 3 e)

2x 3

2x 2 – 6x – 12x 2 + 18x

3 2 f ) 3x +36x +23x 6x + 6x

127

EXERCISES AND PROBLEMS 39

Interpret, describe and express 37

The total charge, T, without VAT, for an electricity bill is calculated with the following formula:

There are five ponds in a field. If C is the amount of water that a pond will contain after m minutes, match each pond with the correct expression.

T = F + (RCU – RPR) · P F 8 Fixed costs and meter rental (€)

pond M contains 4 500 litres of water and a tap is turned on, filling it at 4 litres per minute.

RCU 8 Current reading (kWh)

pond N contains 4 500 litres of water and is connected to a water pump which extracts 4 litres per minute.

RPR 8 Previous reading (kWh)

pond P contains 4 cubic metres of water and is connected to a water pipe that fills it at 4.5 cubic metres per hour.

a) Write the updated formula, if the fixed costs are €8.50 and the kilowatt-hour cost is €0.80.

P 8 Price of one kWh (€/kWh)

b) Which of the following is the updated formula for the total charge, including VAT at 21 %?

pond Q contains 4 cubic metres of water and a water pipe is turned on, extracting 4.5 cubic metres per hour. C = 4 000 + 4 500 · m 60 C = 4 000 – 4 500 · m 60 38

C = 4 500 – 4 · m

8.50 + (R CU – R PR) · 0, 80 + 21 100

T = [8.50 + (RCU – RPR) · 0.80] · 1.21

C = 4 500 + 4 · m

In Marta’s class, the maths grade is calculated according to three different factors: a student’s average test mark (3/4), their workbook (20 %) and special projects (the remainder).

T = 8.50 + (RCU – RPR) · 0.80 + 1.21 40

Last month, an employee of an electricity company read the meter at the Gutiérrez family’s home as 2 457 kWh. This month it read 2 516 kWh. How much did the bill increase by this month?

a) Which of these formulas is used to calculate the grade? Tests (a); Workbook (b); Special proj. (c).

N = 3a + b + c 4 5 20

N = 0.75a + 0.2b + 0.05c

N = 15a + 4b + c 20

N = 75a + 20b + 5c 100

b) Calculate Marta’s and Javier’s marks to two decimal places.

a. test mark

workbook

s. projects

marta

7.25

javier

6.80

c) If the school secretary’s computer system only supports marks with integers, what will their final grades be in mathematics? 128

A plumber wants to calculate an invoice. She charges a fixed price of €25 for attending a call, plus the cost of the materials used, plus €15 per hour for the work. 21 % VAT is also added to all of this. Write the formula for the invoice total (T), based on the hours worked (h), the cost of the materials (M), and the VAT.

Unit 6

Count the number of diagonals in these polygons:

Now, like in the previous activity, look for an expression for calculating the sum, En, of the first n even numbers. 2 + 4 + 6 + 8 + 10 + … 8 En = ? n addends help

— Check that the number of diagonals from a vertex is equal to the number of sides minus three.

• Look at these towers.

— Check that each diagonal intersects two vertices.

Keeping this in mind:

n. of sides

n.o of diagonals

10 20

b) Find the formula for calculating the number of diagonals (D), given the number of sides (n). ‘+’ problems 43

Problem solved

2 + 4 + 6 + 8 + 10

a) Complete this table in your notebook: o

52+ 5

• Try the first cases, compare the results with the previous problems and draw conclusions. E1 = 2 8 1 + 1 E2 = 2 + 4 = 6 8 4 + 2 E3 = 2 + 4 + 6 = 12 8 9 + 3 E4 = 2 + 4 + 6 + 8 = 20 8 16 + 4 45

Write an expression that would allow us to calculate the sum, On, of the first n odd numbers. 1 + 3 + 5 + 7 + 9 + ... 8 On = ? n addends

Find a formula for calculating the sum of the first n natural numbers. help: Look at the five-storey tower and find the general formula for any tower of n storeys. Then, check using examples.

help

• Try the first few cases:

O1 = 1

1+2+3+4+5

O2 = 1 + 3 = 4 O4 = 1 + 3 + 5 + 7 = 16… • Copy and complete the following table: O2

52 + 5 2 2

O3 = 1 + 3 + 5 = 9

1+3+5+7+9

O10 O15

...

Count the number of cards needed to build this three-storey house of cards. a) How many cards would it take to build a 5-storey house of cards? b) And for a 10-story one? c) And for a house of cards of n storeys?

• Look at these shapes. Do they confirm your conclusions?

1 + 3 + 5 + 7 + 9 = 25

52 = 25

129

MATHS WORKSH LEE E AND READ INFÓRMATE LEARN How to generalise

Sometimes numerical sequences are given by their first terms, or by a few particular cases. For example, the number of triangular pieces that make up each of the triangles you see on the right. However, when we look at these first triangles, we immediately ask ourselves: What if the triangle had 20 rows? And what if it had 100? And what if it had…?

…

We can find the solution to these questions in the following table: rows

…

pieces

16 … n2

n rows ↔ n2 pieces 20 rows ↔ 400 pieces

By studying these particular cases, we have found the relationship between the number of rows of any triangle in the sequence and the number of pieces it contains. We have generalised the sequence. Your turn In an exhibition, we presented this hexagonal mosaic which has 3 triangular units per side, and is made of 54 triangular pieces. a) How many pieces do we need to build a mosaic of the same shape, but with 20-unit sides? b) In general, how many pieces do we need to construct a hexagon with n units per side? help:

Think about the example above and look at the hexagons below.

9 side pieces

INVESTIGATE

…

6.1

6.4

6.9

Jump and take Objective: Remove all the counters from the board, except one. Rules: In each move, one counter jumps over another and lands in the next box, which must be empty. The counter that is jumped over is taken and is removed from the board. • Find a code that can be used to easily express these moves. 130

Remember that you can find academic and professional guidance related to this content at anayaeducacion.es.

Unit 6

ENTRÉNATEMAKES PRACTICE RESOLVIENDO PERFECT!OTROS PROBLEMAS Cristian puts several marbles in a box and performs the following experiment:

SELF-ASSESSMENT

He tosses a coin and bets half the number of marbles in the box on the outcome. • If it lands on heads, he adds that number of marbles to the box. • If it lands on tails, he removes it. Given that statistically, when we toss a coin, it lands half the time on heads and half on tails, answer the questions: a) Suppose there are 16 marbles in the box. What outcomes might you expect after tossing the coin four times? What is the probability of getting each outcome? b) Suppose there are x marbles in the box. What outcomes might you expect after tossing the coin twice? c) Do you think the game is advantageous for the box? Why? anayaeducacion.es Answer key and interactive self-assessment.

1 Complete the blank boxes in your notebook,

following the logic of the table. 1

6 Simplify these expressions:

a) 2(x 2 – 5) – 3(2x – 1)

b) x (x – 1) + x 2(3x – 2) 7 Look at these polynomials and calculate:

2 If x is any number, express the following in algebraic

A = 3x 3 + 5x 2 – 6x + 8

language.

a) A + B

a) Double the number.

b) A – B

b) The number after double the number. c) Double the next number. 3 What is the coefficient and what is the degree of these

a) 5x b) a2b2

c) – 2 xy 2 3

4 Simplify these expressions:

+ 2 + 6x – x –

3x 2

c) 3x – (x 2 + 3 – 4x) + 4x 2 a) – 1 x 2(–5x) 5

Commitment

b) 6x 4 : 2x 3

b) (1 + 2x)2

c) (x – 3) · (x + 3)

10 Take out the common factor.

a) 3a 2 + 6a

5 Calculate and simplify.

a) (x – 3)2

b) 4x 3 + 6x 2 – 2x

11 Simplify.

a) 2x + 4 + x – 6 b) 5x 2

8 Calculate the product (2x – 1) · (x 3 + 3x – 6). 9 Calculate.

d) Three times half the number. monomials?

B = x 3 – 5x 2 + 1

3a 2 + 6a

x2 – 9 b) 2 x – 6x + 9

12 Which of the following formulas can we use to

calculate the sum, S, of the first n multiples of 5?

c) da + ab n : 2a 9 9

2 5 (n 2 + n) 5n 2 + n c) a) 4n + n b) 5 2 2

Watch the video for target 16.2. Think of something you can do to contribute to achieve that goal. Make a commitment to put your idea into practice.

131

EQUATIONS Reading and listening

Some people think of Diophantus (3rd century CE) as the ‘father of algebra’ because of his major contribution to improving algebraic terminology and the ingenious methods he devised for solving some types of equations. Diophantus’ writings contain problems solved without relying on geometric representation. However, Diophantus did not discover a general method for solving equations. His work was translated into Arabic in the 10th century and into Latin in the 16th century, and he had a great influence on future mathematicians. In fact, most historians consider Al-Khwarizmi (9th century) to be the true father of algebra. Even though the problems he solved were much simpler in his book Al-jabr (algebra) he demonstrated how equations could be solved using clear, logical and systematic argumentation. As a result, his work spread widely in subsequent eras. 1 In groups, discuss who you think is ‘the father of algebra’: Diophantus

or Al-Kwarizmi.

2 Write a text comparing both mathematicians. Use comparative

adjectives.

Rhetorical algebra and symbolic algebra 3 Observe and solve the equations using trial and error. How many sheep

and how many goats are there in the herd? ordinary language

algebraic language

There are 36 sheep and goats in a herd, and I count as many horns as there are sheep.

Sheep 8 x Horns 8 x Goats 8 x 2 Sheep + Goats = 36 x + x = 36 2

132

Algebraic statements and expressions Read each statement and its translation into graphic and symbolic language. • A jar of jam weighs 150 g more than a packet of biscuits. x

(x + 150)

• The packet and the jar weigh 800 g. (x + 150) x

800 g

8 x + (x + 150) = 800

• If we swap the jar with another packet and 150 g, they weight 800 g in total. 2x

150 g

800 g

8 2x + 150 = 800

–150 –150

• If we remove 150 g from each side of the scales, we see that two packets weigh 650 g. 2x

650 g

8 2x = 650

: 2

• Therefore, a packet of biscuits weighs 325 g. x

325 g

8 x = 325

• And the jar of jam weighs 450 g. 325 + 150 = 475 g

ANK ANK B E G A U LANG LANGUAGE B ANK ANK GE BANK B B E E G G A A U U GUA K LANG ANG N L A L AN GE BANK 133 ANK GE BANK B B E E G G A A U U G A LAN LANG LANGUA LANGU

EQUATIONS: MEANING AND USE An equation expresses a relationship between quantities whose value we do not know at the beginning through algebraic equalities. The quantities are represented by letters. Examples

• In a henhouse, there are a total of 28 beaks, combs and feet:

x chickens

Hens 8 x Beaks 8 x Combs 8 x Feet 8 2x

Equation 8 x + x + 2x = 28

• A shop window is one metre longer than it is wide, and it has an area of 3.75 m2: Width 8 x Length 8 x + 1

• A goods lorry takes an hour longer on its outbound journey, travelling at 60 km/h, than on the return leg, travelling empty at 80 km/h:

x+1

60 km/h

80 km/h

Time (hours)

Outbound journey 8 x Return leg 8 x – 1

Distance (km)

Outbound 8 60 · x Return 8 80 · (x – 1)

outbound journey = return leg 60x = 80(x – 1) x 3

2 . x x

Equation 8 x · (x + 1) = 3.75

+ 30

Equation 8 60 · x = 80 · (x – 1)

• Double a number is equal to a third of the number plus thirty units. _ The number 8 x b Double the number 8 2x b` Equation 8 2x = x + 30 3 A third of the number 8 x bb 3a Equations give us a code for expressing relationships between numbers in algebraic language. This allows us to calculate them mathematically. As we will see later, they are a very powerful tool for solving problems. But first, you must learn how to solve equations.

ãã What is involved in solving an equation? Solving an equation involves finding the value, or values, that the letters take to make the equality true. Example

18 3

2 . 18 18

In the equation in the last example, 2x = x + 30, the equality is only true when 3 the value of x = 18.

+ 30

2x = x + 30 4 2 · 18 = 18 + 30 3 3 x = 18 36 36

We can therefore say that the solution of the equation is x = 18. 134

Unit 7

ãã Equations with infinite solutions and equations with no solution • In the equation 0 · x = 0, any value of x makes the equality true. 0 · x = 0 8 Has infinite solutions • In the equation 0 · x = k, with k ≠ 0, no value of x can make the equality true. 0 · x = k 8 Has no solution

ãã Solve equations using what you already know Before learning any specific technique, remember that you can solve a lot of equations with what you already know, or by trial and error. Examples

• Explain the following process: 7x – 10 = 5 8

• Use trial and error: Z ] For x = 1 8 ] ]] x 3 + 5 = 10 [ For x = 2 8 x+8 ]… ] ] For x = 5 8 \

anayaeducacion.es Solve problems.

Let’s practise! 1

Which equation matches each statement? a) A third of a number is equal to a quarter of the number plus 20 units. (Number 8 x) b) Andrés is triple the age of his sister, and both their ages add up to 20. (Andrés 8 x years) c) A rectangle is 3 metres longer than it is wide and its perimeter is 30 metres. (Width 8 x metres) d) I paid €30 for 3 drawing pads and a box of paints, but the box cost double the price of a pad. (Pad 8 x Euros) e) A cyclist travelled the distance from A to B at a speed of 15 km/h and a person walking at a speed of 5 km/h took 1 hour longer. (Cyclist 8 x hours) f ) A cricket moves forward one metre less than a grasshopper with each jump. In 15 jumps, the cricket travels as far as the grasshopper does in 5 jumps. (Grasshopper 8 x metres) x + x = 20 2x + 2(x + 3) = 30 15(x – 1) = 5x 3 x = x + 20 3 4

3x + 2x = 30

25 = 5 8 7x – 10 = 25 8 7x = 35 8 x = 5

15x = 5(x + 1)

1 3 + 5 = 6 ≠ 10 1+8 9 3 2 + 5 = 13 ≠ 10 2+8 10

x=5 is a solution

5 3 + 5 = 130 = 10 5+8 13

2 Solve these equations in the order that they appear:

a) 3x = 21

b) 3x – 1 = 20

c) 3x – 1 = 4 5

d) 3x – 1 = 2 5

3 Solve these equations with what you know:

a) 6x = 24 c) 2x – 4 = 6 e) x = 9 3 g) x + 1 = 2 3 2 i) x + 1 = 26

b) x + 3 = 10 d) 2(x + 1) = 12 f ) x – 2 = 5 2 h) 7 = 1 x +1 j ) 3x + 1 = 5

4 Solve the following equations by trial and error:

a) x 2 + 2x + 1 = 4 b) x 2 – 5x + 6 = 0 c) x + 8 = 3 d) x 3 – x = 0 x 4 5 Demonstrate that the values of x are solutions to the accompanying equations. a) x 2 – 4x – 5 = 0 b) x 3 = 2x 2 + 3x x = –1 x = 5

x = 0 x = –1 x = 3

135

EQUATIONS: ELEMENTS AND NAMES • The sides of an equation are each of the expressions that appear on both sides of the equals sign. • The terms are the addends that form the sides.

left side

right side

3x + 1 = 9 – x

terms • The unknowns are the letters that appear in the equation. Examples

x x

3x + 1 = 9 – x 8 Equation with one unknown, x. 5x + 3y = y + 2 8 Equation with two unknowns, x and y. • The solutions are the values that the letters must represent to make the equality true. Example

x x x

3x + 1 = 9 – x Solutions: x = 2 is a solution, because 3 · 2 + 1 = 9 – 2. x = 1 is not a solution, because 3 · 1 + 1 ≠ 9 – 1. • The degree of an equation is the largest degree of the monomials that make up the sides once the equation has been reduced. Examples

3x + 1 = 9 – x 8 First-degree equation. x 2 – 3x + 1 = 2x – 5 8 Second-degree equation. • Equivalent equations are two equations which have the same unknowns and the same solutions. Example

3x + 1 = 9 – x 3 They are equivalent. 4x = 8 They both have the solution x = 2. Let’s practise! 1 True or false?

a) x 2 + 6x – x 2 = 7x – 1 is a second-degree equation. b) 2x + x · y = 6 is a second-degree equation.

2 Copy these equations and match them to their solution.

a) 4x + 4 = 5 c) x 2 – 3 = 2x

c) The terms of an equation are the addends that make up the sides. d) An equation can have more than two sides.

b) 4x – 3 = x + 3 d) 3x = x + 1 –1

1 4

1 2

3 Group the equivalent equations together.

e) All first-degree equations are equivalent.

a) 4x = 20

b) 3x – 1 = 8

f ) The equation x + 1 = 5 is equivalent to the equation x + 2 = 6.

c) 5x – 4 = x

d) 3x = 9

e) 4x – 5 = 15

f ) 4x – 4 = 0 anayaeducacion.es Equivalent equations test.

136

Unit 7

3 anayaeducacion.es Practise basic techniques for solving equations.

TRANSPOSING TERMS Transposing or changing the position of the terms is a basic technique that can transform equations into other, simpler equivalent equations. We do this by moving the terms from one side of the equality to the other. The transposition of terms is based on the following principal:

When we add, subtract, multiply or divide the same number in both sides of an equation, we get another equivalent equation. ➜➜ first case: x + a = b

What was being added on one side will now be subtracted on the other side.

x +3=4 3 We subtract 3 from both sides. x =4 –3 ➜➜ second case: x – a = b

What was being subtracted on one side will now be added on the other side.

x x

x – 2=3 3 We add 2 to both sides. x =3+2 ➜➜ third case: a · x = b

What one side was being multiplied by, the other side will now be divided by.

Do not forget x+a=b 8 x=b–a x–a=b 8 x=b+a a·x=b 8 x= b a x =b 8 x=b·a a

2x = 6 4 x = 6 We divide both sides by 2. 2 ➜➜ fourth case: x = b a What one side was being divided by, the other side will now be multiplied by. x =4 4 We multiply both sides by 3. 3 x =4·3 Transposing the terms of an equation allows us to solve the equation. In other words, we isolate the unknown on one side of the equality.

Let’s practise! 1 Isolate the unknown using the instruction given in each case and calculate the solution.

a) x + 7 = 10 8 Subtract 7 from each side. b) x + 8 = 5 8 Subtract 8 from each side. c) x – 4 = 6 8 Add 4 to each side. d) x – 1 = 4 8 Add 1 to each side. e) 6x = 12 8 Divide each side by 6. f ) –2x = 8 8 Divide each side by –2. x = 1 8 Multiply each side by 7. g) x = 5 8 Multiply each side by 3. h) 3 –7 2 Find the unknown and calculate the solution.

a) x + 2 = 5

b) x + 3 = 2

c) x + 8 = 0

d) x – 1 = 5

f ) 3x = 6

g) 5x = 15

h) –5x = 10

i) x = –1 4

e) x – 5 = 1 j) 5x = –10 –2

anayaeducacion.es Solving equations by transposing terms. 137

SOLVING SIMPLE EQUATIONS To solve an equation, we transform it, step by step, making simpler equivalents with each step until we have isolated the unknown. We use two methods for transforming an equation into a simpler equivalent equation: • Reducing its sides (R). • Transposing the terms (T).

Study the following examples and solve the equations that follow. The answers are on the right, so you can check your work.

Remember • The equation 0 · x = 0 has infinite solutions. • The equation 0 · x = k with k ≠ 0, cannot be solved.

Solutions

Example 1

t r

Example 2

3x – 8 = 4

3x = 4 + 8

3x = 12 t x = 12 8 x = 4 3

➜➜ practise

1 2

2 1

3 0

4 –1

5 –2

6 –5

7 6

8 – 4

9 2

Solutions 10 3

11 1

12 –1

13 2/3

14 –1/3

15 –1/2

16 I. S.(*)

17 N.S.(**)

18 I. S.(*)

3 7x – 5 = –5 6 2x – 1 = –11

7 7 = x + 1

8 0 = 3x + 12

9 8 = 5x – 2

Example 4

4 – 3x = 10

–3x = 10 – 4

6x – 3 = 1 6x = 1 + 3

6x = 4 t x = 4 8 x = 2 6 3

➜➜ practise

10 5 – x = 2 13 3x – 1 = 1

11 6 – 2x = 4 14 4 = 3x + 5

12 4 – 5x = 9 15 5 = 4x + 7

16 0x + 2 = 2

17 1 + 0x = 4

18 –5 = 0x – 5

Example 5

t r

Example 6

5x + 1 = 3x + 7

5x – 3x = 7 – 1

2x = 6 t x = 6 8 x = 3 2

4x – 1 = 6x – 9 4x – 6x = –9 + 1

–2x = –8 t x = –8 8 x = 4 –2

➜➜ practise

19 1

20 3

21 –2

22 –4

23 1/2

24 1

25 1/5

26 I. S.

27 N.S.

138

2 = 4x 2 =x8x= 1 2 4

2 2x + 4 = 6 5 2 + 3x = –4

–3x = 6 t x = 6 8 x = –2 –3

(*) 8 I. S. (infinite solutions). (**) 8 N.S. (no solution).

Solutions

7 – 5 = 4x

1 3x – 2 = 4 4 x + 5 = 4

Example 3

7 = 4x + 5

19 8x – 4 = 5 – x 22 3x + 15 + 2x = –5 25 4 = x + 5 – 6x

20 5x + 3 = x + 15 23 5 + 2x + 1 = 7 26 7x + 2 – 7x = 3 – 1

21 10 + x = 2 – 3x 24 5 – 1 = x + 5 – 2x 27 5x + 3 – 5x = 7

anayaeducacion.es Guided activities to practise solving equations.

Unit 7

ãã Same destination, different routes As the equations become more complicated, we can use different methods to solve them. Any of them are valid, as long as you do it correctly. Below, you can see an example solved in two different ways. Example 7

Eliminate the brackets.

2x – 1 – 5x = 2 + 3x + 1 –3x – 1 = 3 + 3x t t

–3x – 3x = 3 + 1

–6x = 4 t x = 4 –6

The unknown in the side with the positive coefficient.

–1 – 3 = 3x + 3x –4 = 6x –4 = x 6

x = –2 3 ➜➜ practise Solutions 28 3

29 2

30 2

31 3

32 2/5

33 –1

34 1

35 –1/2

36 3/5

37 –5

38 I.S.

39 N.S.

28 2x – 1 = x + 2

29 3x + 2 = x + 6

30 2x + 1 = 5x – 5

31 1 – x = 4 – 2x

32 3 + 7x = 2x + 5 34 6x – 2 + x = 2x + 3

33 x – 6 = 5x – 2 35 7 + 3x – 3 = 4x + 5 + x

36 8x + 3 – 5x = 7 – 2x – 1 38 7x – 4 – 3x = 2 + 4x – 6

37 4x – 1 – 7x = 8 – x + 1 39 2 + 3x – 5 = 4x – 2 – x

ãã Equations with brackets When an equation contains brackets, we start by eliminating them and reducing the equation. Example 8 Remove the brackets

5x – 2(2x – 2) = 8 – (3 + 2x) 5x – 4x + 4 = 8 – 3 – 2x x + 4 = 5 – 2x x + 2x = 5 – 4

➜➜ practise Solutions 40 8

41 0

42 2

43 1/2

44 3/4

45 –1

46 2/3

47 1/6

48 –2

49 1

50 I.S.

51 N.S.

3x = 1 8 x = 1 3

40 x – 7 = 6 – (x – 3) 42 1 – (3x – 9) = 5x – 4x + 2

41 x – (1 – 3x) = 8x – 1 43 13x – 15 – 6x = 1 – (7x + 9)

44 7x – (4 + 2x) = 1 + (x – 2) 46 1 – 2(2x – 1) = 5x – (5 – 3x)

45 2(3x – 1) – 5x = 5 – (3x + 11) 47 7 – (2x + 9) = 11x – 5(1 – x)

48 4(5x – 3) – 7x = 3(6x – 4) + 10 50 16x – 7(x + 1) = 2 – 9(1 – x)

49 4 – 7(2x – 3) = 3x – 4(3x – 5) 51 6 – (8x + 1) = 4x – 3(2 + 4x)

anayaeducacion.es Solving equations.

139

EQUATIONS WITH DENOMINATORS When there are denominators in the terms of an equation, we change the equation into an equivalent one that does not have denominators. To do this, we multiply the two sides of the equation by a multiple of all the denominators. The most suitable multiple is the lowest, in other words the lowest common multiple of the denominators. Example

A similar method • Reduce to a common denominator x +2 = 1+ x 4 5 1 10 Common denominator 8 20

_ x + 2 = 1+ x bb LCM (4, 5, 10) = 20 4 5 10 ` 20 · d x + 2 n = 20 · b1 + x l b We multiply both sides by 20. 4 5 10 a 20x + 20 $ 2 = 20 + 20x When we remove the brackets and simplify 4 5 10 4 the equation, the denominators disappear. 5x + 8 = 20 + 2x _ 5x – 2x = 20 – 8 b b From here, we continue in the way we already know. 3x = 12 ` x = 12 8 x = 4 bb 3 a

5x + 8 = 20 + 2x 20 20 20 20 • Remove the denominators: 5x + 8 = 20 + 2x

anayaeducacion.es Help solving equations with denominators.

To eliminate the denominators in an equation, we multiply both sides by the lowest common multiple of all of them.

eas

id Consolidating

1 Copy, multiply to eliminate the denominators, and solve.

a) 4 – 2x = x + 2 6 Multiply both sides by 3. 3 3 6 x 12 – = x + d 3 3 12 – x = x +

b) 5x – 1 = x – 3 6 Multiply both sides by 12. 3 4 6 60x – = d – d 3 4 6 x –

= 4x –

Let’s practise! 1 Multiply by the number indicated and solve.

2 Solve.

a) 1 + 2x = 1 – 2x 6 Multiply both sides by 5. 5 5 b) x = 1 + 2x 6 Multiply both sides by 15. 3 15 5 c) 7x + 1 = 2 + x 6 Multiply both sides by 10. 5 10 d) 7x – 1 = x 6 Multiply both sides by 18. 9 6 3

3x – 1 = 5x a) 3x + 5 = 2x – 1 b) 2 2 4 6 6 c) 3x + 2x + x = 1 d) 3x – 1 = 3x 4 5 10 2 5 5 x –5=x – e) x + 1 = x + 1 f ) 2 3 3 4 2 6 3 g) x – 3x + 1 = 4x – x 4 10 5 2

Solutions: a) –1/3

Solutions: a) 11 b)10 c) 4/5 d) –1/3 e) –1/2 f) 5 g) 2

b) –1

c) 2

d) 3/8

–1 –1 2 x +1 5

anayaeducacion.es Solving equations with denominators. 140

Unit 7

THE GENERAL METHOD FOR SOLVING FIRST-DEGREE EQUATIONS To solve first-degree equations, it is helpful to organise the calculations according to the steps that we can see in the following example: Example

Z ] ] Remove the brackets. ÄÄÄÄÄÄÄ8 [ ]] \ • Second step: • First step:

Remove the denominators. To do this, we multiply both sides by 12. • Third step: anayaeducacion.es Practise solving different first-degree equations.

Isolate the unknown, reducing and transposing terms.

ÄÄÄÄÄ8

x – 1 = 1– 4 3 x – 1 = 1– 4 3

1 dx – 1 n 2 3 x +1 2 6

12x – 12 = 12 – 12x + 12 3 2 6 4 3x – 4 = 12 – 6x + 2

Z ] 3x + 6x = 12 + 2 + 4 ÄÄÄÄÄ8 ] 9x = 18 [ ]] x = 18 8 x = 2 9 \

Let’s practise! 1 Solve these equations:

3 Find the value of x that makes the equalities true.

a) 3 (1 – x) + 2 = 3x 2 b) 1 – 2x = x – 2dx – 1 n 7 3

a) 2d x + 1 n = 3 d 2x – 1 n + 1 2 3 3 2

c) x – x = 1 dx – 3 n + x 2 3 6 2

c) 5 – 2b x + 1l = x + 3 b x – 1l 5 10 2

b) 5x – d 2x + x n = 1 d9x – 1 n 3 2 3 2

d) 2x – 1 = 1 dx – 7 n 3 2 3 3

d) 3d x – 1 n + x = 5d x – 1 n 10 4 4 10

2 Solve the following equations:

4 Solve the following equations:

a) 1 b x + 1l – 2x = 1 dx – 1 n 2 2 3 6 2

x x a) 2b + l – 3x = 3 d 1 + 2x n – 1 3 5 3 5 10 b) 1 – 2 d x – 1 n = x + 3d 2 – x n 4 5 2 5 2

b) 1 (2x – 3) + 1 = 1 (x – 5) – x 2 3 c) 2d 4x – 7 n + x = 1 – x 3 9 6

c) 1 dx – 1 n – 1 b x – 1l = x + 1 3 2 2 6 4 3

d) 5d x + 1 n – 1 = x – 2b1 – x l 2 6 3 3

d) x – 3d x + 1 n = 1 (4x – 6) 5 3 10

1 Solutions: a) 7/9

b) –7/15

c) 1/4

d) –5/6

3 Solutions: a) 7/6

2 Solutions: a) 1

b) –7/10

c) 3/2

d) –3

4 Solutions: a) 0

b) –1/5 b) –1/2

c) 3 c) I. S.

d) 5 d) N. S.

anayaeducacion.es Practise solving first-degree equations. 141

SOLVING PROBLEMS WITH EQUATIONS With the information we are given in a statement about a problem, we find known elements (data) and unknown elements (unknowns). If we can make an algebraic code from all these elements, and relate them with an equality, we make an equation. By solving the equation and studying the solutions in relation to the statement, we solve the problem. In the next few pages, we will see several examples of the process we need to follow. Example 1

When I climb the stairs at home two at a time, I take four more steps than my sister, who plays basketball and can climb them three at a time. How many steps are there on the stairs? a) Identify the parts of the problem. Express the unknowns algebraically. • Number of steps ÄÄÄÄÄ8 x • Two at a time Ä8 x 2 • Three at a time ÄÄÄ8 x 3 b) Relate the known and unknown elements with an equation. stairs climbed = stairs climbed + 4 two at a time three at a time x = x +4 2 3

c) Solve the equation. x = x + 4 8 3x = 2x + 24 8 3x – 2x = 24 8 x = 24 2 3 d) Use the formula to check the solution of the equation. Solution: There are 24 stairs. Check:

24 = 24 + 4 8 12 = 8 + 4 3 2

eas

id Consolidating

Think about the help given for the following problems and then solve them. 1 If we add a fifth of a number to half of the same

number, it is the same as if we subtracted three from the original number. What number is it? • The number 8 x • Half 8 x 2 x • A fifth 8 • Minus three 8 x – 3 5 half the n.º +

142

a fifth of the n.º

the n.º = minus three

2 When we split my class into groups of three, we

end up with two more groups than if we split it into groups of four. How many of us are there in my class? • Students in my class 8 x • Groups of three 8 x 3 x • Groups of four 8 4 groups groups of three – 2 = of four

Unit 7

Example 2

A hotel puts in an order with a bakery for croissants, at €1.40/unit, and ensaimadas, at €1.80/unit. On the receipt there are 50 units altogether and the total is €78. How many of each item did they order? a) Identify the elements of the problem, expressing the unknown elements algebraically. amount

€ 1.40/unit

€1.80/unit

anayaeducacion.es Practise solving problems similar to example 2.

cost

(€)

croissants

1.40 . x

ensaimadas

50 – x

1.80 . (50 – x)

b) Use an equation to equate the known and unknown elements. cost of cost of croissants + ensaimadas = €78 1.40x + 1.80 (50 – x) = 78 c) Solve the equation. 1.40x + 1.80 (50 – x) = 78 8 1.40x + 1.80 · 50 – 1.80x = 78 8 8 90 – 0.4x = 78 8 12 = 0.4x 8 x = 30 d) Express the solution within the context of the original problem. Solution: The order was for: Croissants 8 30

Ensaimadas 8 50 – 30 = 20

Check: 1.4 · 30 + 1.8 · 20 = 42 + 36 = 78 eas

id Consolidating

Think about the help given for the following problems and then solve them. 3

Target 2.3. A farmer loads a van with 35 boxes. Some of the boxes contain potatoes and weigh 15 kg and the rest contain turnips and weigh 10 kg. The boxes weigh 455 kg in total. How many boxes are there of each vegetable? n.o of boxes potatoes turnips

weight

(kg)

15 . x

35 – x

10 . (35 – x)

weight x boxes + weight (35 – x) = 455 kg of potatoes boxes turnips 4 Rosa is 25 years younger than her father, Juan, and

26 years older than her son Alberto. Their three ages add up to 98. What is each one’s age? • Rosa’s age 8 x • Juan’s age 8 x + 25 • Alberto’s age 8 x – 26 rosa’s + juan’s + alberto’s = 98 years age age age

5 One kilo of apples costs €0.50 more than one kilo of

oranges. Marta bought three kilos of oranges and one kilo of apples for €5.30. How much are the oranges? And how much are the apples? • 1 kg oranges 8 x • 3 kg oranges 8 3x • 1 kg apples 8 x + 0.50 cost of 3 kg + cost of 1 kg = €5.30 of oranges of apples

6 A group of friends have gone to a sandwich bar to

have a snack. A baguette costs one Euro more than a sandwich. They pay 11 Euros for three sandwiches and two baguettes. How much does a sandwich cost? What about a baguette? 3x

2(x + 1) +

= €11

143

7 SOLVING PROBLEMS WITH EQUATIONS Example 3 anayaeducacion.es Practise solving problems similar to example 3.

Three farmers receive €100 000 in exchange for some of their land, which is going to be used to build a motorway. How should they split the money, given that the first farmer has lost double the land of the second, and the second has lost three times as much land as the third? a) Identify the elements of the problem, expressing the unknown elements algebraically. • The third farmer receives ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ8 x • The second should receive three times as much as the third Ä8 3x • The first should receive twice as much as the second ÄÄÄÄ8 2 · 3x = 6x b) Use an equation to equate the known and unknown elements. 3rd farmer’s compensation +

2nd farmer’s compensation

1st farmer’s = 100 000 + compensation

x + 3x + 6x = 100 000 c) Solve the equation. x + 3x + 6x = 100 000 8 10x = 100 000 8 x = 10 000 d) Check that the solution is correct by putting it into the original equation. Solution: The third farmer will receive 8 x = €10 000 The second farmer will receive 8 3x = €30 000 The first farmer will receive 8 6x = €60 000 Check: 10 000 + 3 · 10 000 + 6 · 10 000 = 100 000 Let’s practise! 1 Divide €99 between three people so that the first

receives half as much as the second, and the second receives three times as much as the third. • Third person 8 x • Second person 8 3x • First person 8 3x 2

2 The employees of a garage split a bonus of €5 000

between them. The manager gets double what the mechanics get, and the mechanics get €200 more than the apprentices. How much will each of them get if we know that as well as the manager, there are three mechanics and five apprentices. • Apprentice 8 x • Mechanic 8 x + 200 • Manager8 2 · (x + 200) Think before you start: five apprentices will receive 5x and three mechanics will receive...

144

3 Sara and Jorge split the money their grandmother gives

them equally. Then, Jorge says to Sara: if I give you back the €10 I owe you, I’ll only have half as much as you. How much did their grandmother give them? • Grandmother’s money (€) 8 x • If Jorge gives Sara €10, he will have 8 x – 10 2 • If Sara gets €10 more, she will have 8 x + 10 2

500 litres of diesel have been divided equally between two barrels. How many litres need to be moved from one to the other so that the second barrel has three times more than the first? x litres

3·

= 250 – x

250 + x

Unit 7

Example 4

10 L

A restaurant buys a batch of olive oil for €3.40/L and another of sunflower oil for €1.30/L. How much sunflower oil would they have to mix with each 10-litre jug of olive oil for a litre of the mixture to be worth €2.50?

a) The data: amount

0 €3.4

0/L €1.3

(x + 10) L

(L )

price

(€/L)

cost

(€)

olive oil

3.40

3.40 · 10

sunflower oil

1.30

1.30 · x

mixture

10 + x

2.50

2.50 . (10 + x)

b) The equation: cost of 10 L of olive oil

of x L cost of (10 + x) L + of cost sunflower oil = of mixture

3.40 · 10 + 1.30 · x = 2.50 · (10 + x)

mixture

c) Solution of the equation: 3.40 · 10 + 1.30 · x = 2.50 · (10 + x) 8 34 + 1.3x = 25 + 2.5x 8 8 34 – 25 = 2.5x – 1.3x 8 9 = 1.2x 8 x = 9 8 x = 7.5 1.2

0 €2.5

d) Solution: For the mixture to be worth €2.50/L, they would have to add 7.5 litres of sunflower oil to every 10 litres of olive oil. Check: 8 3.40 · 10 =

€34.00

Cost of 7.5 litres of sunflower oil 8 1.30 · 7.5 =

+ €9.75

Cost of 10 litres of olive oil

Cost of the mixture (17.5 litres) 8 2.50 · 17.5 = €43.75 Let’s practise! 5 A shopkeeper has two types of coffee: a high-quality

coffee that costs €12.70/kg and a low-quality coffee that costs €7.80/kg. How many kilos of high-quality coffee should she mix with 100 kilos of low-quality coffee in order to make a mixture of medium-quality coffee that costs €9.90/kg? kilos

price

(€/kg)

cost

(€)

7 Martina mixed some red and yellow paint together to

get 40 litres of orange paint. A litre of red paint costs €3.40, and a litre of yellow paint costs €2.60. If the orange paint costs €2.95 per litre, how many litres of each type did she use? litres red

high quality

12.70

12.70 · x

yellow

40 – x

low quality

100

7.80

7.80 · 100

mixture

x + 100

9.90

9.90 (x + 100)

6 A goldsmith has two gold alloys, a 96 % noble metal

alloy and a 75 % one.

How many grams of the first alloy would she have to mix with 100 grams of the second to increase its purity to 82 %?

price

(€/L)

cost

(€)

2.95

8 A businesswoman buys 100 shirts at €24 each and

puts them on sale for 30 % more. However, after selling some of the shirts, she puts the rest on sale. The discounted price is only 10 % more than the price she bought them for. By the time she has sold all the shirts, she has made an overall profit of 25 %. How many shirts did she sell at the higher price?

145

7 SOLVING PROBLEMS WITH EQUATIONS Example 5

A rectangular farm is 80 metres longer than it is wide and the fence around the whole farm is 560 metres. What are the dimensions of the farm? a) The data:

x + 80

Shorter side (width) 8 x

x x Longer side (length) 8 x + 80 Length of the fence 8 560 m x + 80 b) The equation: 2x + 2 · (x + 80) = 560 c) Solving the equation: 2x + 2 · (x + 80) = 560 2x + 2x + 160 = 560 4x = 560 – 160 8 4x = 400 8 x = 100 d) Solution: Shorter side 8 100 m

Longer side 8 100 + 80 = 180 m 100 100 + 80 The farm land measures 180 m long by 100 m wide. Check: Length of the fence: 2 · 100 + 2 · 180 = 200 + 360 = 560 metres Let’s practise! 9 We need 150 metres of fencing to go around a

rectangular farm that is twice as long as it is wide. What are the dimensions of the farm? 2x

11 In an isosceles triangle, the base is 11 cm longer than

the two equal sides, and the perimeter is 65 cm. Find the length of each side. x

x x + 11

10 In a scalene triangle, the medium side measures 7 cm

more than the shortest side and 5 cm less than the longest side. What is the length of each side, if the perimeter is 52 cm? x–7

x x+5

146

12 In a trapezium whose perimeter measures 51 cm, the

longer base is twice as long as the shorter one, which is 3 cm shorter than each of the non-parallel sides. Find the length of each side. x x+3

x+3 2x

Unit 7

SECOND-DEGREE EQUATIONS An equation is second-degree if, after being simplified, it meets these conditions: • One of its terms is a second-degree monomial. • It does not contain terms of degree greater than two. For example, the following is a second-degree equation: 2x + 5x 2 – 1 = 3 + 4x 2 + 5x

second-degree monomials

The equation above can be simplified as follows: 2x + 5x 2 – 1 – 3 – 4x 2 – 5x = 0 8 x 2 – 3x – 4 = 0

general form of the equation Any second-degree equation with one unknown can be shown in the following general form:

Remember

ax 2 + bx + c = 0 where a ≠ 0, b and c are known coefficients.

(x – 3) · (x – 2) × – + x 2 – x 2 –

x x 2x 3x 5x

– 3 – 2 + 6

For example: equation

general form 8

+ 6

coefficients

5x 2 = 45

8 8

5x 2

(x – 3) · (x – 2) = 0 x 2 – 5x + 6 = 0

+ 0x – 45 = 0

a = 5, b = 0, c = – 45

a = 1, b = –5, c = 6

ãã Solving second-degree equations Generally, a second-degree equation has two different solutions. However, some have one double solution and others do not have any solution.

Keep in mind Solutions to second-degree equations are also called roots. x =4 x 2 – 3x – 4 = 0 ) x = –1 The roots of the equations are x = 4 y x = –1.

Example

The equation x 2 – 3x – 4 = 0 has two solutions: )

x =4 x = –1

For x = 4 8 42 – 3 · 4 – 4 = 16 – 12 – 4 = 0 For x = –1 8 (–1)2 – 3 · (–1) – 4 = 1 + 3 – 4 = 0

Let’s practise! 1 Say which of these equations are second-degree and

write them using the general form: a) x 2

b) x 2

= 5

c) 2x(x – 1) = 4 e) 7x 2

– 4x =

x 2

+3=

x 2

d) x(x – 3) = + 2

f ) 5x + 6 –

x 2

–1 7x 3

g) 3x 2 + 9 – 3x 2 = x h) x 3 + 2x = x(x + 3)

2 Match each equation with its solution:

a) x 2 = 25

b) x 2 = 9

c) x 2 + x – 6 = 0

d) x 2 – 7x + 10 = 0

e) x 2 + 3x – 10 = 0

f ) x 2 – 5x + 6 = 0

–5

–3

147

SOLVING SECOND-DEGREE EQUATIONS ãã The equation x 2 = k In order to solve the equation x 2 = k, we look for the numbers whose square is k. In other words, we look for the square root of k. x 2 = k 8 x = ± k If k is a positive number, there are two opposite solutions; if k is negative, it cannot be solved. Examples

• x 2 = 36 8 x = ± 36 8 x = )

+6 –6

• x 2 + 25 = 0 8 x 2 = –25 8 x = ± –25 It cannot be solved, because there is no square root of –25.

Keep in mind Often, you have to give the solutions in an approximate form. For example: 3x 2 – 15 = 0 8 x 2 = 15 3 8 x=± 5 5 8 {“…“«\\|£} x=)

≈ 2.24 ≈ –2.24

ãã The equation ax 2 + c = 0 It is similar to the case above. ax 2 + c = 0 8 ax 2 = –c 8 x 2 = –c 8 x = ± –c a a If the radicand is positive, there are two solutions; if it is negative, the equation cannot be solved. Examples

+3 • 2x 2 – 18 = 0 8 x 2 = 18 8 x = ± 9 8 x = ) –3 2 • 5x 2 + 20 = 0 8 x 2 = –20 8 x = ± – 4 . It cannot be solved. 5

ãã The equation ax 2 + bx = 0 Here, we want to take out common factor from the left side. ax 2 + bx = 0 8 x · (ax + b) = 0 If a product is equal to zero, one of the factors has to be zero, which gives us two options: x = 0 6Ä first solution x · (ax + b) = 0 * ax + b = 0 8 x = –b 6Ä second solution a Examples

• x 2 – 5x = 0 8 x · (x – 5) = 0 ) anayaeducacion.es Help for solving incomplete second-degree equations.

148

• 5x 2 – 2x = 0 8 x · (5x – 2) = 0

x =0 x – 5=0 8 x =5 x =0

* 5x – 2 = 0

8 x= 2 5

Unit 7

ãã The equation ax 2 + bx + c = 0 anayaeducacion.es Practise using the formula for second-degree equations.

The process for solving a complete second-degree equation is long and complicated, so we use a formula that gives us the unknown. This formula will help you find the answers quickly and easily. formula: ax 2 + bx + c = 0 8 x =

We will study this formula in more detail in later years. For the moment, it is important that you learn it and remember how to use it as shown in the following examples.

Double solution (x – 5)2 = 0 a =1 2 x – 10x + 25 = 0 * b = –10 c = 25 x=

– (–10) ± (–10) 2 – 4 · 1 · 25 = 2 ·1 10 + 0 = 5 2 10 – 0 = 5 2

= 10 ± 0 = 2 ·1

–b ± b 2 – 4ac 2a

Examples

• 5x 2 – 7x + 2 = 0 8 a = 5; b = –7; c = 2 2 x = – (–7) ± (–7) – 4 · 5 · 2 = 7 ± 9 = 2·5 10

• 5x 2 + 6x + 2 = 0 8 a = 5; b = 6; c = 2

7 + 3 =1 10 7–3 = 2 10 5

2 x = –6 ± 6 – 4 · 5 · 2 = –6 ± –4 2·5 10 The equation cannot be solved, because the square root of – 4 does not exist.

eas

id Consolidating

1 Copy, complete the coefficients and calculate the solutions.

a) x 2

a=1 + 2x – 3 = 0 8 * b = 2 c = –3

b) 5x 2

a=5 – 3x – 2 = 0 8 * b = –3 c = –2

2 2 x = –d ! d – 4 $ d $ (–3) = … x = – (–d) ! (–d) – 4 $ d $ (–2) = … 2$d 2$d

Let’s practise! 1 Solve the following equations:

a) x 2

= 81 2 d) x – 9 = 0

b) x 2

= 25 2 e) x + 6 = 10 2 5x 2 = 2 g) x = 7 h) 7 8 5

3 Calculate the solutions using the formula.

c) 5x 2

= 20 2 f ) 4x + 1 = 2 2 i ) 2x – 1 = 0 9 50

2 Simplify, take out the common factor and solve.

a) x 2

– 4x = 0 2 d) x + x = 0

b) x 2

c) x 2

+ 2x = 0 –x=0 2 2 e) 3x – 2x = 0 f ) 5x + x = 0 2 2 x2 = x g) x = x h) i ) x + x = 5x 2 3 3 3 4 6

a) x 2 – 6x + 8 = 0 c) x 2 + x – 12 = 0 e) 2x 2 – 7x + 6 = 0 g) x 2 + 6x + 9 = 0

b) x 2 – 6x + 5 = 0 d) x 2 + 7x + 10 = 0 f ) x 2 – 2x + 1 = 0 h) x 2 – 3x + 3 = 0

4 Reduce and solve.

a) x 2 – 3x – 5 = 2x + 9 b) 6x 2 – 5(x – 1) = x(x + 1) + 4 c) x (x + 1) – 1 = x – 4 6 2 149

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ROBLEM P D N A S E IS C R EXE

Simple equations

Solve the equation: 3x – 1 = x – x + 1 5 2

Solve mentally. a) x + 4 = 5

b) x – 3 = 6

c) 7 + x = 10

d) 7 – x = 5

e) 9 = 15 – x

f ) 2 – x = 9

c) 6x – 1 + x = 4 – 5x + 3

10 $ d 3x – 1n = 10 $ dx – x + 1 n 5 2 6x – 10 = 10x – 5(x + 1) 6x – 10 = 10x – 5x – 5 6x – 10 = 5x – 5 6x – 5x = 10 – 5 8 x = 5

d) x + 2x + 3x – 5 = 4x – 9

Solve. a) 2x – 5 + 3x + 1 = 3x – 2 b) x + 7 = 12x – 3 – 8x + 1

e) 5x + 4 – 6x = 7 – x – 3 3

Remove the brackets and solve the equations. a) 6(x + 1) – 4x = 5x – 9 b) 18x – 13 = 8 – 4(3x – 1) c) 3x + 5(2x – 1) = 8 – 3(4 – 5x) d) 5 – (4x + 6) = 3x + (7 – 4x) e) x – 7(2x + 1) = 2(6 – 5x) – 13

f ) 11 – 5(3x + 2) + 7x = 1 – 8x g) 13x – 5(x + 2) = 4(2x – 1) + 7 First-degree equations with denominators 4

Problem solved

3x · d1 + 2 + 1 n = 3x · 2 x 3 3x 3x + 6 + x = 2 8 4x = 2 – 6 4x = – 4 8 x = – 4 8 x = –1 4

b) 5 (2x – 1) – x = x 6 6

c) x – 1 = 2 dx – 4 n d) x – 1 = 1 (2x – 5) 5 5 3 6 150 150

Solve the equations. a) 3x – 1 – 2x + 1 = 7x – 13 4 5 20 b) 2 + 2 (x + 1) = x – 2x + 3 5 5 ( x – ) 3 1 c) 2 (1 – 3x) + = 5 (1 – x) 3 4 12

Solve the equation: 1 + 2 + 1 = 2 x 3 3x

Remove the brackets and the denominators, then solve the equations. a) 2x – 5 = 1 (x – 3) 2 2

b) 1 – 1 – x = x + 1 3 2 d) x + 2 – 3x = x + 1 5 2 f ) 3x – 1 = x – x + 1 5 2 h) 1 – x – x – 1 = 3x – 1 3 12 4

d) 3 d x – 1 + 1n + x = 3 dx – 2 n 5 3 4 3

Remove the denominators and solve the equations. a) 5x + 1 = 5 + x 6 3 b) 3x – 1 = x – 7x – 1 5 4 10 5 c) x + 4 – x = 1 – 7x 6 10 3 15 d) 7x – 1 – x = x + 5x + 1 8 8 4 e) x + 1 – x = 5 + x – 2 2 6 3 6 6 3

Remove the denominators and solve the equations. a) 1 – x + 1 = 2x – 1 3 3 c) 3x – 1 – 1 = 2x – 2 2 e) 2x + x – 3 = x – 3 2 4 g) x + 3 – x – 6 = 1 5 7

f ) 4x + 2 + 7x = 10x + 3 + x

Problem solved

Solve like you did in the previous problem solved. a) 2 + 1 = 5 + 1 b) 1 + 1 = 1 + 1 2x 5 5x 2 x 2 3x 3 1 +3= c) 1 – 2 = 1 + 1 d) 2x 9x 3x x – 1 2 2 (x – 1) Multiply by 6x, 10x, 18x and 2(x – 1), respectively.

Unit 7

Second-degree equations 11

Look, explain and solve. a) 5x 2 = 45 b) 12x 2 = 3 c) x(x – 3) = 0 d) (x + 5)x = 0 e) x(3x – 1) = 0 f ) 3x(5x + 2) = 0 2 g ) x – 7x = 0 h) x 2 + 4x = 0 i) 3x 2 = 2x j ) 5x 2 = x 2 – 2x Solve using the formula. a) x 2 – 10x + 21 = 0 b) x 2 + 2x – 3 = 0 c) x 2 + 9x + 40 = 0 d) 5x 2 + 14x – 3 = 0 e) 15x 2 – 16x + 4 = 0 f ) 14x 2 + 5x – 1 = 0 g) x 2 – 10x + 25 = 0 h) 9x 2 + 6x + 1 = 0 i ) 6x 2 – 5x + 2 = 0 j ) 6x 2 – x – 5 = 0

The sum of two numbers is 167, and the difference between them is 19. What are these numbers?

Calculate the natural number which when added to the following number gives 157. • The number 8 x • The number after it 8 x + 1

The sum of three consecutive numbers is 135. What are these numbers?

Teresa is seven years older than her brother Antonio and two years younger than her sister Blanca. How old are they if their three ages add up to 34 years? Antonio 8 x – 7; Teresa 8 x ; Blanca 8 x + 2

a) x 2 – 1 = 1 b x – 1l 4 5 4

An ensaimada, a pastry from Mallorca, costs 10 cents more than a croissant. Three croissants and four ensaimadas cost 6 Euros. How much does each cost?

c) x d 1 – x n = 2 – x dx – 5 n 3 2 2 3

Nicolás bought two pairs of trousers and three shirts for €161 in the sales. If we know that a pair of trousers cost twice as much as a t-shirt, what was the price of each item?

Clara runs with strides of 80 cm and her sister, Sara, runs with strides of 75 cm. It takes Sara 50 strides more to do a lap of the athletics track than Clara. How many strides does each of them take to do one lap? How long is the track?

Reduce to the general form and apply the formula.

b) x dx – 1 n = x d x + 5 n 2 2 6 2 3

2 2 d) x + x = 2x – 5 – 1 2 3

Solving problems with first-degree equations 14

First calculate mentally and then use an equation to help. a) When you add 12 to a number, you get 25. What number is it? b) If you subtract 10 from a number, you get 20. What is the number? c) Adding a number, x, to the following number, x + 1, gives us 13. What are these numbers? d) There are 29 students in total in my class, but there are three more boys than girls. How many boys and how many girls are in the class?

When we double a number and add three to it, the result is the same as tripling it and subtracting five. What is the number?

If we multiply a number by 5, we get the same answer as when we add 12 to it. What is that number?

In a supermarket cash register there are 1 140 Euros organised in 5, 10, 20 and 50 Euro notes. Given that: — There are double the number of €5 notes as there are €10 notes. — There are the same number of €10 notes as there are €20 notes. — There are six more €20 notes than there are €50 notes. How many notes of each are in the cash register?

A gardener plants half of his garden with melons, one third with tomatoes, and the rest, 200 m2, with potatoes. What is the area of her garden? • Area of garden 8 x • Melons 8 x/2 • Tomatoes 8 x/3 • Potatoes 8 200 m2 151

EXERCISES AND PROBLEMS 26

When he walks, Adrián has strides of 0.80 m, but when he runs his strides are 1.25 m. If he does a lap of the athletics track while walking, it takes 180 more strides than if he does it running.

Adela is six times older than her grandson Juan, but in eight years she will only be four times older. How old are Juan and Adela? age today

How many strides does he take in each case? How long is the track? 27

A bakery makes two batches of muffins, with the same number in each batch. The first batch is packed into bags of 10 muffins and the second into bags of 12. How many muffins are there in each batch if 5 more bags are used for the first batch than the second? • N.º of muffins per batch 8 x

8 years

juan

x+8

adela

6x + 8

age of adela in eight years 32

age in

=4.

age of juan in eight years

A cyclist rides up a hill at 15 km/h, and then descends along the same path at 35 km/h. If the route lasted 30 minutes, how long did he take going up?

• N.º of bags of 10 muffins 8 x 10 • N.º of bags of 12 muffins 8 x 12 28

• Time of ascent 8 x (hours)

A water tap fills a tank in 30 minutes. Another tap, which is 2 litres per minute slower than the first, fills a second tank in 20 minutes.

• Time of descent 8 1/2 – x (hours) • Distance travelled going up 8 15x • Distance travelled going down 8 35d 1 – x n 2

If the second tank has the capacity to hold 200 litres of water more than the first, what is the flow rate of each tap? 29

Problem solved

distanced travelled = distanced travelled ascending descending 33

Joaquín is 14 years old, his sister is 16 and his mother is 42. In how many years will the sum of the children’s ages equal the age of their mother? age today

age in x years

joaquín

14 + x

sister

16 + x

mother

42 + x

In x years, the following will occur: joaquín’s + age

sister’s age

= mother’s age

• Time until they meet 8 x (hours) • Distance travelled by the first 8 24x • Distance travelled by the second 8 16x distanced travelled + distanced travelled = 30 by the first by the second 34

Two trains are at the stations of two cities that are 132 km apart. Both start at the same time, on parallel routes, towards the other city. If the first travels at 70 km/h, and the second at 95 km/h, how long will it take for them to pass each other?

(14 + x) + (16 + x) = 42 + x 2x + 30 = 42 + x 8 x = 12 Solution: In 12 years. 30

A father is 38 years old, and his son is 11. In how many years will the father be double the age of the son?

152

Two cyclists start their routes at the same time: one, from A to B, at a speed of 24 km/h, and the other from B to A, at 16 km/h. If the distance between A and B is 30 km, when will they meet?

A cyclist leaves a town, by road, at a speed of 22 km/h. An hour and a half later a motorist leaves to try and find the cyclist, travelling at a speed of 55km/h. How long will it take him to reach the cyclist?

Unit 7

A cyclist is travelling by road at a speed of 18 km/h for 20 minutes. At what speed does she have to travel for the next 10 minutes so that the average speed is 20 km/h for the 30 minutes she rode the bike?

A customer paid €66 for a garment that was reduced by 12 %. What was the original price? • Original price 8 x • Discount 8 12x 100 • Equation 8 x – 12x = 66 100 €6

Laura bought a skirt and a blouse for €59. The skirt cost €16 more than the blouse. There was a 20 % discount on the skirt, and only 15 % on the blouse. How much did each garment cost? cost without discount

cost with discount

skirt

0.80 . x

blouse

x – 16

0.85 . (x – 16)

discounted + discounted = €59 skirt blouse 39

A cheese manufacturer has mixed a certain amount of cow’s milk that cost €0.50/L with another quantity of sheep’s mik, which costs €0.80/L. The result is 300 litres of mixture with an average price of €0.70/L. How many litres of each type of milk was used? quantity

(L )

price

(€/L )

cost

(€)

cow

0.50

0.5 . x

sheep

300 – x

0.80

0.8 . (300 – x)

mixture

300

0.70

0.7 · 300

cost of + cost of = cost of cow’s milk sheep’s milk mixture

The amplitude of one of the angles of a triangle is 13 degrees greater and 18 degrees less, respectively, than the amplitude of the other two angles. Calculate the measurements of each angle. x + 18 x

Problem solved

Part of a rectangular plot has been allocated for roads. They have allocated 10 m from the length of the plot and 10 m from its width, making a total area of 800 m2. If the resulting rectangle is 50 meters long, how wide is it? a) We use a diagram to help us: 10 x 50

In order to mark a rectangular area, which is twice as long as it is wide, we need 84 m of tape. What are the dimensions of the marked section?

b) We represent the data and the unknown algebraically: • Original area 8 60 · (x + 10) • New area 8 50 · x 60 · (x + 10) – 50 · x • Lost area 800 m2 c) We write the equation and solve it: 60 · (x + 10) – 50x = 800 60x + 600 – 50x = 800 8 10x = 200 8 x = 20 d) Solution: The plot is 20 metres wide. 44

A square wooden board is cut along two lines, which are perpendicular to one another and parallel to the sides of the board. This creates two new planks of wood, which are 20 cm and 15 cm wide, respectively. What were the dimensions of the board if the planks have a total area of 0.32 m2? x

A company buys a tank of concentrated juice for €0.35/L. 35 litres of water is added to dilute it. This reduces the cost per litre by 7 cents. How much juice was in the tank before diluting it?

x – 13

x 15 cm 20 cm 153

EXERCISES AND PROBLEMS Analyse and explain 45

Study the solutions below the problem and explain how the equation was formed in each one. 16 m Calculate the perimeter of this property, given that it x has an area of 930 square 3x metres. 960 m2 40 m

x 3x

Calculate the integer for which the product of the number before it and the number after it is 11 units greater than the result of multiplying the number itself by four. Find all the solutions. (x + 1) · (x – 1) = 4x + 11 2 x – 1 = 4x + 11 8 x 2 – 4x – 12 = 0

47 40 m

40 m

40 · 3x – 24 · x = 960 120x – 24x = 960 8 96x = 960 8 x = 10 m Perimeter = 140 m Solution C

b) The sum of the squares of two consecutive numbers is 5. What are these numbers? x 2 + (x + 1)2 = 5 48

Solution D

49 x 3x

24 m 2x

40 m

3x + x · 16 + 24 + 40 · 2x = 960 2 2 2x · 16 + 32 · 2x = 960 8 96x = 960 8 x = 10 m Perimeter = 140 m 154

Calculate mentally, then with an equation.

a) What number, when multiplied by the number following it, makes 12? x · (x + 1) = 12

960 – 16 · x = 40 · (3x – x) 960 – 16x = 40 · 2x 8 960 = 80x + 16x 960 = 96x 8 x = 10 m Perimeter = 140 m 16 m

x= 4!8 2

12 = 6 2

–4 = –2 2 Solution: There are two integers that satisfy the conditions of the problem: 6 and –2. Check: (6 + 1) · (6 – 1) 8 35 6 4 · 6 + 11 (–2 + 1) · (–2 – 1) 8 3 6 4 · (–2) + 11

24 m

16 m

Problem solved

96x = 960 8 x = 960 8 x = 10 m 96 Perimeter = 40 + 20 + 24 + 10 + 16 + 30 = 140 m 16 m

2 x = 4 ! 4 – 4 $ 1 $ (–12) = 4 ! 64 2$1 2

Solution A 16 · 3x + (40 – 16) · (3x – x) = 960 48x + 24 · 2x = 960 8 48x + 48x = 960

Solution B

Solving problems with second-degree equations

If you increase a number by three units and multiply it by the same number decreased by three units, the product is 55. What number is it? (x + 3) · (x – 3) = 55 Multiplying my age by the age of my sister, who is 5 years younger than I am, gives three times the age of my mother, who had me when she was 28 years old. How old am I? • My age 8 x • My sister’s age 8 x – 5 • My mother’s age 8 x + 28

This morning, six more cars left the parking garage on my block than entered it. If the product of the ones leaving and the ones entering is equal to the total number of wheels of all the cars, how many cars left and how many cars entered the garage?

Unit 7

‘+’ problems

Problem solved

Calculate the dimensions of the rectangle if it is 7 cm longer than it is wide and its area is 120 cm 2. x · (x + 7) = 120 120 cm2

x 2 + 7x – 120 = 0 x+7 2 x = –7 ± 7 – 4 · 1 · (–120) = –7 ± 23 2 ·1 2 Solution:

8 + 7 = 15 m

A rectangular farm occupies an area of 6 hectares and has a perimeter of 1 000 metres. Calculate the dimensions of the farm. Remember: 1 ha = 10 000

6 ha

We fill a pond with water using two water pipes: A and B. Using just pipe A, we can fill the pond in 3 hours. Using both pipes, we can fill it in 2 hours. How long will it take to fill if we only use pipe B? A

A+B

time

fraction filled in an hour

1 3

1 x

1 2

fraction fraction fraction filled by A in + filled by B in = filled by both 1 hour 1 hour in 1 hour 1+1=1 3 x 2 Multiply both sides by 6x: 6x + 6x = 6x 8 2x + 6 = 3x 8 x = 6 3 x 2 Solution: It will take 6 hours to fill the pond using only pipe B. 56

500 – x

In a right-angled triangle, the hypotenuse is one metre longer than the longer leg, which is one metre longer than the shorter leg. Find the length of each side.

A water tank has two taps. If we only turn on the first tap, the tank fills in 8 hours and if we turn on both taps, it fills in 3 hours. How long does it take to fill if we only turn on the second tap?

x–2

Problem solved

8 –15

8 . 15 = 120 m2

The members of my team give a gift to the coach which costs €80. It is a little expensive, but if there were two more of us, each of us would pay two Euros less. How many players are on the team? • N.º of players on the team 8 x • Each player must pay 8 80 x

x–1

In a trapezium with an axis of symmetry, the larger base is three times the length of the smaller base, which is 3 cm longer than the height. If the area of the trapezium is 216 cm2, calculate the perimeter. x

• If there were two more, each would pay 8

what each – 2 = what each one would pay one pays if there were two more 58

x–3 3x

80 x +2

A car travels from A to B at the same time that a lorry goes from B to A. They take 2 hours to pass each other at an intermediate point in the journey. How long has it taken the car to complete the full journey if the lorry has taken 5 hours? 155

MATHS WORKSH LEE E AND READ INFÓRMATE LEARN

History of the x In writings by Arab mathematicians prior to the 15th century there are references to equations. They called the unknown element the thing, because they had no other name for it. ‘How much is the thing if when increased by five it is double the thing minus seven?’ The word thing, in Arabic is pronounced xay, so this is the expression translators used. Later, it was abbreviated to its first letter, x.

x + 5 = 2x - 7 Equations: degree and difficulty Finding formulas to solve first- and second-degree equations is not very difficult for mathematicians. Those problems were solved in the Middle Ages. But this was not the case with problems that had a higher degree than two. In fact, during the Renaissance, Italian mathematicians of the 16th century, who were considered the most advanced at the time, worked hard in this area. Some of the most notable of these were Niccolo Fontana (1501-1557), nicknamed Tartaglia, and Girolamo Cardano (1501-1576), who were rivals and became embroiled* in bitter disputes (Tartaglia accused Cardano of stealing his work). F ocu s on Eng lish Mathematicians continued working on this problem, but by the end of the 16th embroil: to become involved century they still had not been able to find formulas for solving equations with a higher in conflicts, arguments or degree than four. Then, in the 19th century the Norwegian genius Niels Henrik Abel difficult situations. (1802-1829, he died aged 27!) demonstrated that these formulas generally do not exist.

INVESTIGATE You can do it! However, we can solve some equations of the third degree or higher. For example, look at these equations: x 3 – 2x 2 – 5x + 6 = 0 (x – 1) · (x + 2) · (x – 3) = 0 With what you have studied so far, you can only solve the first one by using trial and error. However, you can see that the solutions to the second one are: x=1 x = –2 x=3 And these are also the solutions to the first equation. Check it and verify that by multiplying the brackets you get the same equation. • Are you ready to solve these three now? (x + 1) · (x + 3) · (2x – 1) = 0 x 3 – 9x = 0 x 3 – 9x 2 = 0 • Do you know how to make an equation with x = 5, x = 1 and x = –2 as its solutions? 5

156

Remember that you can find academic and professional guidance related to this content at anayaeducacion.es.

Unit 7

ENTRÉNATEMAKES PRACTICE RESOLVIENDO PERFECT!OTROS PROBLEMAS Draw a diagram, do the maths, use trial and error • When the clock on the tower strikes the hour, there is a gap of one and a half seconds between chimes. How long will it take to strike midday? • It takes an aizkolari wood cutter a quarter of an hour to chop a tree trunk into three pieces. How long will it take him to chop another tree trunk as thick as the first into six pieces? • A farmer sells her tomatoes to a wholesaler. The wholesaler sells to a vendor, making a 20 % profit. The vendor sells them to a warehouse, making 20 %. The warehouse sells them to a shop, and the shop, to the public, each of them also making 20 %. By what percentage did the tomatoes increase in price, from what the farmer charged to when the public finally bought them? • Place the numbers from 1 to 9 in each circle, so that all sides of the triangle add up to 23. There are two solutions.

SELF-ASSESSMENT

anayaeducacion.es Answer key and interactive self-assessment.

1 Indicate which of the following values is a solution of

the equation:

x=1

–1 = x –1 5

x=2

x=4

x=9

x= –1 2

2 Solve.

a) 7x – 3 – 2x = 6 + 3x + 1

b) x – x + 1 = x + 3 – 2 5 2

2x – 4 d x – 1 n = 2 c) x – 1 = 5x – 3 d) 3 5 6 15 2 8 4 4 Solve.

– 5 = 70

c) x 2 – 2x – 3 = 0

Commitment

b) 6x 2

3x – 8 = x – 3 2 x

6 Ramon paid €7.80 for three kilos of pears and two

kilos of apples. If we know that the price of the pears is one and a half times the price of a kilo of apples, what is the cost of each item? garden with cabbages and 3/10 with carrots. If there is still 110 m2 free, what size is the vegetable garden?

3 Solve.

a) 3a 2

equation.

7 A gardener has planted 1/3 of the area of his vegetable

b) 1 – 4x – 6 = x – 3(2x – 1) a) 3 (2x + 4) = x + 19 4

5 Use the general formula to find the solution to the

– 3x = x

8 If we shorten a square by 4 cm on one side and 3 cm

on the other, its area is reduced by half. What length were the sides of the square? x

x–4

d) 8x 2 – 6x + 1 = 0 Watch the video for target 4.a. Think of something you can do to contribute to achieving that goal. Make a commitment to put your idea into practice.

x–3

157

Systems of Equations Reading and listening

Mathematicians from Mesopotamia and Babylonia already knew about equations in the 17th century BCE. Writings translated from a clay tablet in Babylonia already contained systems of equations related to simple everyday problems, such as: 2 widths + length = 10 hands 3 lengths + width = 15 hands They solved these problems using a different method each time; no general system was developed. The Egyptians and Greeks used a similar approach. In the 2nd century BCE, the Chinese made great advances. They learnt to solve systems of equations with ease, but this knowledge did not reach the West for many centuries. The appearance of symbolic algebra in Europe from the 15th century laid the foundations for new methods of solving equations, as well as sets of equations with several unknowns (systems of equations). 1 Write 3 sentences in the first conditional: If + present simple + will +

+ infinitive

Example: If we translate writings, more people will understand them. 2 Write 3 questions using question words (who, what, when, where, why,

how) about the main idea in the text. Answer your questions in complete sentences. Then, write a short text with that information.

Use your ingenuity 3 How many red balls will it take to balance the scales?

158

4 How much did each of them put in?

Yes, but you put in €250 more than I did.

We have enough for the trip now. Between the two of us we have €1 450.

5 Find a value for a and another for b that make both equations true:

2a + b = 10 3b + a = 15 Are these equations related to the Babylonian problem?

Think and solve 6 Look at the images and explain the process shown in algebraic language.

€7.00

€1.20

× 2

€?

€2.90 €5.80 3a + 2b = 7.00 ⎯8 3a + 2b = 7.00 × 2 a + b = 2.90 ⎯8 2a + 2b = 5.80 a + 0 = 1.20 8 a = 1.20 8

)

b=?

How much does one croissant cost? And an ensaimada? 7 Solve using a similar a process to the activity above.

= =

+ €0.80

ANK ANK B E G A U How much does a watermelon cost? LANG LANGUAGE B ANK ANK GE BANK B B E E G G A A U U GUA K LANG ANG N L A L AN GE BANK 159 ANK GE BANK B B E E G G A A U U G A LAN LANG LANGUA LANGU

FIRST-DEGREE EQUATIONS WITH TWO UNKNOWNS A first-degree equation with two unknowns expresses the relationship between two unknown values. Example

We do not know the length of one engine, x, or one coach, y, for each of the following trains. Remember In the equation:

2x + 3y = 86 8 y = 86 – 2x 3 for any value we give x, we get a corresponding value for y. x

–2 30

1 28

2.5 27

16 18

... ...

In other words, it has infinite solutions.

y 86 m

But in both cases we can say that: 2x + 3y = 86. Also note that the lengths of x and y are different for each train. In other words, the equality may be true for different values of x and y. For example: x = 16 4 8 2 · 16 + 3 · 18 = 86 y = 18

x = 19 4 8 2 · 19 + 3 · 16 = 86 y = 16

We say that these pairs of values (x, y) are solutions of the equation, and we can see that there is more than one solution. But in fact, if we give x any value, we get a corresponding value for y. In other words, the equation has infinite solutions. General Form All linear equations can be written in this way:

So, if we want to find the length of each engine and each coach, we need more information.

ax + by = c where a, b, and c are known values.

• First-degree equations with two unknows are called linear equations. • A solution of a linear equation is one pair of values that makes the equality

true. • A linear equation has infinite solutions.

Let’s practise! 1 Find which of the following pairs of values are

solutions for the equation 3x – 4y = 8: a) *

x =4 x =3 x =0 x =1 * * * b) c) d) y =1 y =2 y = –2 y = –1

2 Find three different solutions for the following

equation:

2x – y = 5

3 Copy and complete the table of solutions for the

equation 3x + y = 12 into your notebook. x

3 9

5 0

–1

–3 18

4 Reduce these equations to the general form.

a) 2x – 5 = y

b) x – 3 = 2(x + y)

c) y = x + 1 2

anayaeducacion.es Practise first-degree equations with one unknown. 160

Unit 8

ãã Graphical representation of a linear equation To find different solutions for a linear equation, we usually isolate one of the unknowns and give values to the other. The values are shown in order in the table below. For example, let’s take the equation that associates the weight of a ball (x) and a die (y) on the scales that appear on the left. Y

3x + y = 45 We isolate y.

45 g

y = 45 – 3x

I f we give different values to x, we get the corresponding values of y. 8x 8y

–5

…

–15 60

…

30 20 10 –20 –10

When we represent the values on the graph, they appear in a straight line.

–10

Each point of the red segment corresponds to one possible solution for the weight of the ball and the die (positive values). • Each linear equation has a straight line associated with it on the graph.

anayaeducacion.es Practise the graphic representation of linear equations.

• Each point on the straight line represents one of the infinite solutions to the

linear equation.

eas

id Consolidating

1 Copy and complete the table for the following equation:

x – 2y – 4 = 0 8 x – 4 = 2y 8 y = x – 4 2 x

– 8 – 6 – 4 – 2

– 6

– 4

…

2 –10

–8

–6

–4

–2

X 10

–2 –4

Draw the graph in your notebook and plot the pairs of values. Make sure that they form a straight line.

–6

Let’s practise! 5 Copy the table for each equation in your notebook

and graph the corresponding straight line.

a) x – y = 0 8 y = x b) x – 2y = 2 8 y = x – 2 2 x

– 6

– 4

–2

…

6 Graph the following equations:

c) y = x + 3 2 7 Write the equation and graph its straight line. a) 2x – y = 1

b) 2x + y = 1

= €8

…

anayaeducacion.es GeoGebra. Graph a linear function. 161

SYSTEMS OF LINEAR EQUATIONS • Two linear equations form a system: *

ax + by = c a'x + b'y = c'

• The solution to a system is the common solution of both equations. Example

The following two equations form a system: *

5 (2, 3) x – 2y = –4

Look at the tables of solutions for each equation:

Solution

x – 2y = – 4 8 y = x + 4 2

3x – y = 3 8 y = 3x – 3 X

–2

5 3x – y = 3

–1

…

–2

…

– 6 –3

…

The solution is the pair of values *

–5

solution to a system: *

3x – y = 3 x – 2y = – 4

x =2 y =3

x =2 which is true for both equations. y =3

In the graph, we can see that both lines go through point (2,3). In other words, the lines intersect at that point. The solution to a system of linear equations is the point of intersection of the lines associated with the equations. ➜➜ special cases

systems with no solution

systems with infinite solutions

The equations are incompatible.

The equations are equivalent.

The straight lines are parallel.

The straight lines overlap.

For example: *

x – 2y = 2 x – 2y = 6

x–y=2

x – 2y = 2

anayaeducacion.es Practise finding the graphical solution of linear equations.

2x – 2y = 4

x – 2y = 6

x– y=2 2x – 2y = 4

Let’s practise! 1 Graph and write the solution.

Z ]] y = 2 + x x+ y=4 x– y=3 2x – 3y – 6 = 0 2 * * a) * b) [ c) d) x x– y=2 2x + y = 0 2x + y + 2 = 0 ]y =4– 2 \ anayaeducacion.es GeoGebra. Graph systems of linear equations.

162

Unit 8

METHODS FOR SOLVING LINEAR SYSTEMS We are now going to learn some techniques for solving systems of equations. They allow us to find, from two equations, another equation with a single unknown. Once we solve that equation, it is easier to find the value of the second unknown.

anayaeducacion.es

ãã Substitution method

• Help for finding the solution using the substitution method. • Practise finding the solution for systems using the substitution method.

First, we isolate one of the unknowns in one of the equations. Then the resulting expression is substituted into the other equation. Problem solved

Solve with the substitution method: *

• For example, we isolate x in the second equation. Then we substitute the resulting expression into the first equation:

3x – y = 3

x=2 y=3

x + 2y = 8

3x – y = 3 x + 2y = 8

3(8 – 2y) – y = 3 3x – y = 3 x + 2y = 8 8 x = 8 – 2y

• Now we have an equation with a single unknown. Let’s solve it: 3(8 – 2y) – y = 3 8 24 – 6y – y = 3 8 7y = 21 8 y = 21 8 y = 3 7 • Now that we know the value of y, we can easily calculate the value of x: x = 8 – 2y 8 x = 8 – 2 · 3 8 x = 2 Solution to the system 8 *

x =2 y =3

eas

id Consolidating

1 Copy, complete and solve the system above, but this time isolate y in the first equation.

3x – y = 3 8 y = ... x + 2y = 8 x + 2 (…) = 8

Solve the resulting equation to get the value of x: x = ... Now that you know the value of x, calculate the value of y, which you have already isolated: y = ... Let’s practise! 1 Solve by substitution and check that your solutions match the solutions shown below.

a) *

x = 2y y = x +1 x + 2y = 11 2x – y = 1 x + 2y = 1 * * * * b) c) d) e) x + 3y = 10 3x – 2y = 7 3x – y = 5 5x – 3y = 0 2x + 3y = 4

Solutions: a) x = 4 y = 2

b) x = 9 y = 10

c) x = 3 y = 4

d) x = 3 y = 5

e) x = 5 y = –2

anayaeducacion.es Solve systems by using the substitution method. 163

3 METHODS FOR SOLVING LINEAR SYSTEMS

ãã Equalisation method We isolate the same unknown in both equations and equalise the resulting expressions. Let’s solve the same exercise from the previous page and compare the difference between both methods. Problem solved

Keep in mind The equalisation method, solving the unknown y: • *

3x – 2y = 3 8 y = 3x – 3 2x + 2y = 8 8 y = 8 – x 2

3x – 3 = 8 – x 2 • 6x – 6 = 8 – x 8 7x = 14 8 8 x=2

• y = 3x – 3 8 y = 3 · 2 – 3 8 8 y=3

Solution 8 *

x =2 y =3

Solve with the equalisation method: *

3x – y = 3 x + 2y = 8

• For example, we can isolate x in both equations and equalise the two resulting equations: Z ]] 3x – y = 3 8 x = 3 + y 3+ y 3 = 8 – 2y [ 3 ] x + 2y = 8 8 x = 8 – 2 y \ • Now we have an equation with a single unknown. Let’s solve it: 3+ y = 8 – 2y 8 3 + y = 24 – 6y 8 7y = 21 8 y = 21 8 y = 3 3 7 • We substitute the value y = 3 into either of the equations from the first step and calculate x:

anayaeducacion.es Help for solving systems with the equalisation method. Practise solving systems with the equalisation method.

x = 8 – 2y 8 x = 8 – 2 . 3 8 x = 2 Solution for the system 8 *

x =2 y =3

eas

id Consolidating

2 Copy and complete to solve the following system by equalising, isolating the unknown y :

4x + y = 1 8 y = – 4x 3x – y = –15 8 y = 3x +

– 4x = 3x +

8 –7x =

8 x = d 8 x = ... –7 y = 1 – 4d 8 y = ...

Let’s practise! 2 Solve by equalisation and check that your solutions match the solutions below.

a) *

y = 3x x + 2y = 3 2x + y + 6 = 0 5x + 2y = 0 2x – y = 3 * * * * b) c) d) e) y = 5x – 4 x – 3y = 8 5x – y + 1 = 0 2x + y = 1 4x – 2y = 7

Solutions: a) x = 2 b) x = 5 c) x = –1 d) x = –2 y = 6 y = –1 y = – 4 y = 5

e) No solution.

anayaeducacion.es Solve systems by using the equalisation method. 164

Unit 8

ãã Reduction method • Once the equations are in the general form, we multiply them by the appro-

anayaeducacion.es

priate numbers so that the coefficients of one of the unknowns are opposite. • By adding the equations together, the unknown disappears.

• Help for solving systems with the reduction method. • Practise solving systems with the elimination method.

Problem solved

Solve with the reduction method: *

Keep in mind Solving the system by reducing x: • *

3x – y = 3 ⎯8 3x – y = 3 × (–3) x + 2y = 8 ⎯8 –3x – 6y = –24

3x – y = 3 + –3x – 6y = –24 – 7y = –21 • y = –21 8 y = 3 –7 • 3x – y = 3 8 3x – 3 = 3 8 8 3x = 6 8 x = 2 Solution 8 *

3x – y = 3 x + 2y = 8

• We multiply the first equation by +2 so that the coefficients of the unknown y are opposite (+2, –2) and we add both equations together: ×2 3x – y = 3 ⎯⎯8 6x – 2y = 6 * x + 2y = 8 ⎯⎯8 + x + 2y = 8 7x + 0y = 14 • We solve the resulting equation: 7x = 14 8 x = 14 8 x = 2 7 • We substitute the value x = 2 into any of the original equations:

x + 2y = 8 8 2 + 2y = 8 8 2y = 6 8 y = 3 Solution to the system 8 *

x =2 y =3

eas

id Consolidating

3 Copy, complete and follow the instructions to solve the following systems by

reduction in your notebook.

a) Add the equations together to eliminate y. *

7x + 2y = 6 7x + 2y = 6 8 + x – 2y = 10 x – 2y = 10

8x + 0y = 7x + 2y = 6 8 7 . 2y = –

8 x = ...

+ 2y = 6 8

8 y = –d 8 y = ... 2

b) Multiply the first equation by 2 and the second equation by 3 to eliminate x. × (–2) 3x – 5y = 5 ⎯⎯8 –6x + dy = –d × 3 2x – 3y = 4 ⎯⎯8 + 6x – dy = d 0x + y = 8 y = ...

2x – 3y = 4 8 2x – 3 . = 4 8 2x = 8 x = d 8 x = ... 2

Let’s practise! 3 Follow the instructions and solve by reduction.

a) *

4x + y = 1 (Multiply the 1st equation by +3) x – 3y = 10

b) *

2x + 3y = 7 (Multiply the 1st equation by 3x – 5y = 1 +5 and the 2nd by +3)

anayaeducacion.es Solve systems by using the reduction method. 165

SOLVING PROBLEMS WITH SYSTEMS OF EQUATIONS Systems of equations are a powerful tool for solving problems. Study these examples. They can serve as a model for solving other similar problems. Example 1

Sara and her younger brother Alberto are on a seesaw. The seesaw is balanced with Alberto carrying a 3-kilo rucksack. Then they both sit on the same side of the seesaw without the rucksack. The seesaw is balanced with their father on the other side. He weights 87 kilos. How much does each child weigh? 3 kg

87 kg

• Identify the elements of the problem and write them algebraically. sara’s weight 8 x

alberto’s weight 8 y

• Express the relationships between these elements with equations.

Solve the system (y + 3) + y = 87

Sara weights 3 kilos more than Alberto. 8 x = y + 3

2y + 3 = 87

Together, Sara and Alberto weigh 87 kilos. 8 x + y = 87

2y = 84

• Solve the system:

y = 42

x=y+3 x = 42 + 3

x = y +3 x + y = 87 8 (y + 3) + y = 87 8 y = 42 8 x = 45

• Express the solution within the context of the problem and check it.

x = 45

Solution: Sara pesa 45 kilos, y Alberto, 42 kilos. Check: 45 = 42 + 3 45 + 42 = 87 eas

id Consolidating

Look and solve. How much does each box weigh? 175 kg

125 kg

x = y + 175 dy = x + d

8y 2 Pepa is 5 years older than her brother Enrique. Their ages add up to 21.

How old is each of them? pepa’s age 8 x

enrique’s age 8 y

3 In a class of 29 students, there are three more girls than boys. How many

boys and girls are there in the class? boys 8 x

166

girls 8 y

pepa’s age = enrique’s age + 5 pepa’s age + enrique’s age = 21 girls = boys + 3 boys + girls = 29

Unit 8

Example 2 anayaeducacion.es Solve a problem involving prices and shopping.

A greengrocer sells pineapples and mangoes at a fixed price per unit. Look at the cost of these two bags of fruit. What is the price of a pineapple? And a mango?

F ocu s on Eng lish We use length (l), height (h) and width (w) to describe three-dimensional objects. The corresponding questions are: How long is it? How tall is it? How wide is it?

€8

€9

• We identify the elements of the problem and write them algebraically. price of a pineapple 8 x

price of a mango 8 y

• We express the relationships between the elements with equations. Two pineapples and a mango cost €8. 8 2x + y = 8 One pineapple and three mangos cost €9. 8 x + 3y = 9 • We solve the system. × 3 2x + y = 8 ⎯8 6x + 3y = 24 * + –x – 3y = –9 x + 3y = 9 × (–1) ⎯8

5x + 0y = 15 8

x=3

y=2

• Express the solution within the context of the problem and check it. Solution: A pineapple costs €3 and a mango costs €2. Check: 2 · 3 + 2 = 8 3+3·2=9 eas

id Consolidating

4 Look at the diagram and solve: What length of shelf does one of these boxes take up when it

is lying on its side? And when it is standing up?

dx + y = d * x + dy = d

y 48 cm

42 cm

5 I have bought three pens and a marker for €6. My friend Rosa has paid €9.25 for two pens

and three markers. How much does one pen cost? What about a marker? 8 €x

dx + y = d * dx + dy = d

8 €y

6 A customer has paid €3.90 for one kilo of oranges and two kilos of apples at the greengrocer.

Another customer has paid €5.70 for three kilos of oranges and one kilo of apples. How much does one kilo of oranges cost? What about one kilo of apples? 1 kg of oranges 8 € x 1 kg of apples 8 € y

7 Last week we paid €10 for two tickets to the movies and a box of popcorn. Today we have

paid €22 for four tickets and 3 boxes of popcorn. How much does one ticket cost? And how much does one box of popcorn cost?

x + dy = d * dx + y = d

dx + y = d * dx + dy = d

167

4 SOLVING PROBLEMS WITH SYSTEMS OF EQUATIONS Example 3

Flora is 17 years older than her nephew Pablo, and in 9 years she will be double his age. How old is Pablo now? What about Flora? I was 17, and in 9 years I’ll be twice as old as you.

How old were you when I was born?

• We identify the elements of the problem and write them algebraically. age now

9 years

flora

x+9

pablo

y+9

• We express the relationships between the elements with equations. Flora is 17 years older than Pablo. ÄÄÄÄ8 x = y + 17 In 9 years, Flora will be twice as old as Pablo. 8 x + 9 = 2(y + 9) • We solve the system.

Ä8 x = y + 17 x = y + 17 * 8 y + 17 = 2y + 9 8 y = 8 8 x = 25 x + 9 = 2 (y + 9) Ä8 x = 2y + 9

• Express the solution in the context of the problem and check it. Solution: Flora is 25 and Pablo is 8. Check: 25 = 8 + 17 25 + 9 = 2(8 + 9)

eas

id Consolidating

8 Complete and solve in your notebook.

a) Cristina is three times as old as her cousin María, but in 10 years she will only be twice as old as María. How old is each cousin? age now

10 years

cristina

x + 10

maría

y + 10

Cristina is three times as old as María. ÄÄÄÄ8 x = y In 10 years, Cristina will be twice as old as María. 8 x + 10 = (y + 10) b) Raphael is six times as old as his granddaughter Adela, but 2 years ago his age was seven times hers. How old is Rafael now? What about Adela? age now

168

two years ago

equations

rafael

x–

x = y

adela

y–

x–

= (y – )

Unit 8

Example 4

If we mix an olive oil that costs €4.30/L with a lower-quality oil that costs €2.80/L we get 500 litres of medium-quality oil that costs €3.40/L. How many litres of each type of olive oil have we used?

x litres €4.30/L

• Identify the elements of the problem and write them algebraically. quantity

y litres €2.80/L

(L )

price

(€/L)

cost

(€)

higher-quality oil

4.30

4.30 · x

lower-quality oil

2.80

2.80 · y

mixture

500

3.40

500 · 3.40

• Use equations to express the relationships between the elements. x+y Litres of the mixture 8 x + y = 500 500 4.3x + 2.8y Cost of the mixture 8 4.3x + 2.8y = 1 700 500 · 3.4 • Solve the system of equations.

500 litres €3.40/L

x + y = 500 8 x = 500 – y 4.3x + 2.8y = 1700 8 4.3 $ (500 – y) + 2.8y = 1700 8 y = 300

x + y = 500 8 x + 300 = 500 8 x = 200 anayaeducacion.es Help with solving problems using systems of equations.

• Express the solution in the context of the problem and check it. Solution: We have used 200 L of higher-quality olive oil and 300 L of lowerquality olive oil. Check: 200 + 300 = 500 200 . 4.3 + 300 . 2.8 = 500 . 3.4

eas

id Consolidating

9 Copy and solve in your notebook.

a) Coffee A is of higher quality and costs €13/kg. Coffee B is of lower quality and costs €8/ kg. How much of each do we need to get 30 kilos of coffee mixture that costs €10/kg? amount

(kg)

price

(€/kg)

cost

(€)

superior coffee

(A)

13x

inferior coffee

(B)

mixture

30 . 10

Amount of coffee A + Amount of coffee B = Amount of mixture 8 x + y = Cost of coffee A + Cost of coffee B = Cost of mixture ÄÄÄÄ8 13x + y = 300 b) What quantities of gold, at €8/gram and silver at €1.70/gram do we need to make 1 kg of an alloy that costs €4.22/gram? quantity

(g)

price

(€/g)

cost

(€)

gold

silver

1.7

alloy

1 000

4.22

…

equations x+y= x + y =

anayaeducacion.es Solve a problem involving mixtures. 169

4 SOLVING PROBLEMS WITH SYSTEMS OF EQUATIONS Example 5

The perimeter of a rectangle is 40 cm and its area is 91 cm 2. How long are the sides of the rectangle? • Identify the elements of the problem and write them algebraically.

length 8 x

perimeter 8 2x + 2y

width 8 y

area 8 x · y

• Use equations to express the relationships between the elements. Perimeter 8 2x + 2y = 40

Remember ax 2

Area 8 x · y = 91 (Note that in this case the equation is not linear.)

+ bx + c = 0

x = –b ±

• Solve the system of equations.

– 4·a ·c 2·a

We have come to a second-degree equation, which we solve in the same way as in the previous unit.

Keep in mind

2 y 2 – 20y + 91 = 0 8 y = 20 ± 20 – 4 · 1 · 91 = 2 ·1

In the equation y 2 – 20y + 91 = 0: a=1

2x + 2y = 40 x + y = 20 8 x = 20 – y 8 * x · y = 91 x · y = 91 8 (20 – y) $ y = 91 8 20y – y 2 = 91

b = –20

c = 91

Solutions to the system

13 7

y = 13 8 x = 20 – 13 8 x = 7 y = 7 8 x = 20 – 7 8 x = 13

Both solutions are equivalent and provide a single solution to the problem. 7 cm

• Express the solution within the context of the problem and check it. Solution: The rectangle is 13 cm long and 7 cm wide. Check: Perimeter 8 13 · 2 + 7 · 2 = 26 + 14 = 40 cm 13 cm

Area 8 13 · 7 = 91 cm2

eas

id Consolidating

10 Copy and complete in your notebook.

a) A rectangle is 7 cm longer than it is wide and has an area of 98 cm2. Calculate the length of the sides. 8 *

x = y +d x $ y = d 8 (y + d) $ y = d 8 …

b) Look at the shape below. When you cut four equal corners from a square, you get another square with sides that are 2 cm shorter and an area that is 24 cm2 smaller. What length were the sides of the original square? What about the new square? x

170

x = y +d 8 * 2 2 x – y = d 8 (y + d) 2 – y 2 = d 8 …

r to choose Remembe rtfolio. po ur yo r fo

resources

from this un

Unit 8

MS ES AND PROBLE

EXERCIS

Linear equations 1

Which of the following are linear equations? Explain your answer. a) 3x + 5y = 1

b) x 2 – y 2 = 6

d) x = 4 – 3y

e) xy – 4 = 8

c) y = 5x – 1 y+1 f ) x – 3 = 2 5

Which of the following pairs of values are solutions to the equation? x=5 x=2 x=1 y=0 y=5 y=2 x + 2y = 5 8 x = –1 x = 0 x=7 y = 3 y = – 4 y = –1

Systems of equations: graphic representation 5

a) * 6

–7

–2

–4

–1

–5

2x – 3y + 15 = 0

2x – 3y + 1 = 0

b) Write a system with this solution: x = 0, y = 5.

c) Write a system with no solution. 7

Solve by graphing and write the solution. x– y=0 x=y * 8 ) x+y = 0 x = –y

X 10

Systems of equations: algebraic solutions 8

D –5

Which of the points are solutions to the equation? Complete in your notebook and answer. Y A

y=x–3 –3

y B

y = 1 – 3x –1

3x – y = 2

a) Write a system with this solution: x = 2, y = 4.

Look at the graph and answer.

B –10

x+ y= 1 x – 2y = 4 * b) x – 2y = –5 3 x – y = –3

x + 2y = 10

Complete the table in your notebook and represent the equation on the graph. y= x+1 8 3

Graph the equations.

a) Which line corresponds to each equation? b) Which point belongs to both lines? c) Which pair of values for (x, y) satisfies both equations at once?

Solve by substitution, appropriate unknown.

isolating

the

a) *

2x + 3y = 8 x – 2y = 7 * b) 5x – y = 3 2x – 3y = 13

c) *

x + 4y = 1 5x – 2y = –5 * d) 2x – y = –7 4x – 3y = 3

most

Solve by using the equalisation method. a) *

y = 3x – 5 x+ y–7=0 * b) y = 5x – 1 x– y+3=0

c) *

x – 3y = 8 5x + 2y = 1 * d) 3x + 5y = 10 7x + 3y = 0

Solve by using reduction. a) *

2x + y = 6 3x + 4y = 1 * b) 5x – y = 1 3x – y = 11

c) *

2x + 3y = 8 3x – 5y = 9 * d) 4x – y = 2 2x – 3y = 5 171

EXERCISES AND PROBLEMS 11

Solve using the most appropriate method. a) *

2y = x + 8 x + y = –4 * b) y = 2x + 10 2x + y = –1

c) * e) * 12

Solve problems using systems of equations 15

x + 2y = –5 3x – y = 1 * d) x – 3y = 5 5x + 2y = 9

The sum of two numbers is 57, and the difference between them is 9. What are the two numbers?

6x – 2y = 0 7x – 5y = 10 * f ) 3x – 5y = 12 2x – 3y = –5

Calculate two numbers knowing that the difference between them is 16, and that double the smaller number is five more than the bigger number.

Alejandro and Palmira have €15 between them. If Alejandro gives Palmira €1.50, she will have twice as much. How much does each of them have?

A 4.80-metre-high bamboo cane is broken by the wind. The top end, which is now pointing towards the ground, is at a height of 60 cm. At what height did the bamboo cane break?

A cyclist rides up a hill and then goes back down the same route. It took him 23 minutes more to go up than to go down, and the entire ride took 87 minutes. How long did it take him to go up the hill? And how long did it take him to go down?

Problem solved

Z y –1 ]] 2 (x – 3) + 1 = 2 Solve this system: [ ] 3 ( x – 2) = 4 ( y + 3 ) + 5 \ • We remove the brackets and the denominators: Z y –1 ]] 4x – 10 = y – 1 2x – 6 + 1 = 2 [ 8 * 3x – 6 = 4y + 17 ] 3x – 6 = 4y + 12 + 5 \ • We write the equations in the general form and solve them by reduction:

At a cafe, we paid €3.80 for two coffees and one fizzy drink the other day. Today we had one coffee and three fizzy drinks, and we paid €4.10. How much is one coffee? How much is one fizzy drink?

×4 4x – y = 9 ⎯⎯8 16x – 4y = 36 * × (–1) 3x – 4y = 23 ⎯⎯8 + – 3x + 4y = –23 13x + 0y = 13 13x = 13 8 x = 1

4x – y = 9 8 4 · 1 – y = 9 8 y = –5

Solution: x = 1, y = –5 13

Solve the following systems:

a) *

2 (3x + y) + x = 4 (x + 1) 6 (x – 2) + y = 2 (y – 1) + 3

Solve. x– y= 2 x– y=3 * 2 a) * 2 b) 2 x – y = 24 x + y2 = 5

172

= €5.90

A fully booked hotel has 62 guests in 35 rooms, some in single rooms, and some in double rooms. How many single and double rooms are there?

A street vendor sells melons and watermelons at a fixed price per unit. Carolina buys 5 melons and 2 watermelons and pays €27 for them. Julian pays €12 for 3 melons and 4 watermelons. How much is one melon? And one watermelon?

A detergent factory packs 550 kg of detergent in 200 bags, some in 2 kg bags, and some in 5 kg bags. How many bags of each type did they use?

5 (2x + 1) = 4 (x – y) – 1 b) * x – y x + 5 = 2 3 Z ] x – 4 – y – 5 =0 ] 3 c) [ 2 y ]] x + = 2x – y \3 4 14

= €3.80

Unit 8

Write a problem that can be solved with the system shown in the picture and solve it. x

3x 25

= €3.90

= €5.70

A shop is selling 100 sets of bed sheets at €70 each. The shop sells a good number of sets and reduces the price of the remaining sets to €50. The sale continues until all are sold. The total income made was €6 600. How many sets have been sold at the original price, and how many at the reduced price?

Double Javier’s age is equal to half of his father’s age. In five years, his father’s age will be three times Javier’s age. How old is each person?

After making improvements to the tracks, a freight train reduces its journey time by 30 minutes, while a high-speed train reduces its journey time by 15 minutes. The ratio of their journey times is now one to seven, whereas before it was one to six.

before

time ratio

1/7

x + 30

y + 15

1/6

How long does each train take to make their journeys? 33

A greengrocer had 80 kg of cherries in his shop. After a few days, he sold most of the cherries. The remaining cherries were not in good condition, so he discarded them. For every kilo of cherries the greengrocer sold, he earned €1, and for every kilo he discarded he lost €2. If the total profit was €56, how many kilos did he sell and how many kilos were discarded?

In a tailoring workshop, they used to take three times longer to make a jacket than to make a pair of trousers. However, after upgrading the machinery, it takes 10 minutes less to make each garment. Now it takes five times longer to make a jacket than a pair of trousers. Time taken for each (min)

On a farm, there are 12 heads and 34 legs between pigs and chickens. How many pigs are there? How many chickens are there?

high-speed

new

freight

Pigs 8 x

Chickens 8 y

Pig legs 8 4x

Chicken legs 8 2y

How long did it take to make a jacket before? And now? And what about a pair of trousers? 34

The length of a rectangle is 7 centimetres shorter than twice its width, and the perimeter is 58 cm. What are the dimensions of the rectangle?

Rosendo has 12 coins in his pocket: some are 20-cent coins and some are 50-cent coins. If he has a total of €3.30, how many coins of each type does he have?

x = 2y – 7 * 2x + 2y = 58

A frog travels the same distance in seven jumps as a grasshopper does in five. If they both take six jumps, the grasshopper will travel 144 cm further than the frog. How far does each travel with each jump?

If Gracia multiplies her age by seven, she gets the age of Concha, her grandmother. But in 11 years she will only have to multiply her age by four to get her grandmother’s age. How old are they? age today gracia concha

x y

11 years

x + 11 y + 11

x 35

To put a fence around a rectangular farm, which is 25 meters longer than it is wide, we needed 210 meters of fencing. Calculate the dimensions of the farm.

The perimeter of an isosceles triangle is 29 cm and the sum of the identical sides is 3 cm longer than the base. Calculate the length of each side. y

y x

x + 2y = 29 2y = x + 3

173

EXERCISES AND PROBLEMS 37

The area of a triangle is 54 m2, and its base is one centimetre longer than two thirds of its height. Calculate the length of the base and the height.

Calculate the length of the sides of the polygon below if the perimeter is 42 cm and the area 73 cm2.

5 cm

A pure olive oil costs €3/litre and an olive pomace oil costs €2/litre. What quantities of each oil are needed to get 600 litres of mixed oil that costs €2.40/litre? Below is a problem solved in two different ways. Indicate the differences and explain the process. A lorry leaves city at 90 km/h. Ten minutes later, a car leaves at 110 km/h. Calculate the time it takes the car to catch up with the lorry and the distance travelled from the starting point. Solution A

A lawn is laid in a square courtyard. There is a path around the lawn with a width of 2 metres and an area of 184 m2. x 2m y

speed (km/h)

time (h)

distance (km)

car

110

lorry

x + 10 60

* y = 90dx + 1 n 8 110x = 90dx + 16 n 8 y = 110x

x = y+d * 2 2 x –y =d

x= 3 h 4 y = 82.5 km

Solution: 45 minutes and 82.5 km. What are the dimensions of the courtyard? What about the lawn? 40

In an isosceles triangle, the two equal sides are both 13 cm long and the height from the third side is 2 cm more than this. Calculate the area.

13 cm

Z ]]x = y + 2 [ 2 y 2 2 ]x + d 2 n = 13 \

Remember the Pythagorean theorem. 41

A TV show gives a €3 000 prize to be shared between two contestants, A and B. The share they get will be proportional to the number of challenges that they overcome. After the challenges are completed, contestant A has overcome five challenges and contestant B has overcome seven. How much does each contestant get?

174

A will get 8 x

B will get 8 y

The prize is proportional to the number of challenges overcome 8 x/5 = y/7

Solution B Car’s distance = lorry’s distance 8 d Car’s time 8 distance/speed = d 110 Lorry’s time 8 distance/speed = d 90 d = d + 1 8 d = 82.5 km 90 110 6 Car’s time 8 d = 82.5 = 3 h = 45 min 110 110 4 Solution: The car takes 45 minutes to catch up with the lorry, and both vehicles have travelled 82.5 km. 44

Two cities, A and B, are 270 km apart. A car sets off from A to B at 110 km/h. At the same time, a lorry sets off from B to A at 70 km/h. How far has each vehicle travelled when they pass each other? The sum of the distances is 270 8 x + y = 270

The time that passes until each vehicle passes each other is the same 8 x/110 = y/70

A pedestrian leaves A to walk to B at a speed of 4 km/h. Simultaneously, a cyclist leaves B to go to A travelling at 17 km/h. If the distance between A and B is 7 km, how long will it take for them to pass each other and how far will they be from A and from B?

Unit 8

‘+’ problems 46

A car and a lorry leave two different cities at the same time, heading towards each other. They pass one another after two hours. When the lorry arrives at its destination, it has been three hours since the car did. How long was each vehicle’s journey?

• Lorry 8 x hours In one hour it covers 1/x of the journey. x = y+3 • Car 8 y hours * In one hour it covers 1/y of the journey. 1 + 1 = 1 x y 2 • Both 8 2 hours In one hour they cover 1/2 of the journey. 47

Water flows into a tank from two taps. When both taps are on, it takes one hour and 12 minutes to fill the tank. How long will it take each tap to fill the tank if one tap takes one hour more than the other? Explain the following system and solve it:

*1+1=5 x = y +1 x

Hint: 1 h 12 min is 72/60 of an hour, which is 6/5 of an hour.

Problem solved

The three digits of a palindromic number add up to 19, and if we swap the hundreds for the tens, it increases by 90. What number is it? First method: trial and error • The palindromic numbers with three digits adding up to 19 are: 919 - 838 - 757 - 676 - 595 • Since there are only a few of them, we just add 90 to each one until we find the solution: 919 + 90 = 1 009 838 + 90 = 928 676 + 90 = 766 Solution: The number we are looking for is 676. Second method: proposing a system of equations • A three-digit palindromic number can be represented algebraically like this: x y x 8 100x + 10y + x • Keeping this in mind, we write the system of equations: x + y + x = 19 4 (100x + 10y + x) + 90 = (100y + 10x + x) x y x y x x • We reduce and solve it: 2x + y = 19 48 x=6 8 y=7 x +1= y

How much does the bottle of orange juice cost? What about the jar of jam? And the box of biscuits?

Solution: x y x 8 6 7 6 The number is 676.

J + B = €5

51 O + B = €4

O + J = €3

Using trial and error and a system of equations, solve a problem like the one above, but with different data. • The sum of the digits is 13.

The area of a trapezium is 204 cm 2. Its height is equal to the smaller base and it is one centimetre longer than half of the bigger base. Calculate the perimeter.

• By exchanging the hundreds for the tens, the number decreases by 180.

x 13 cm

13 cm

x + ]y x + y $ x = 204 — · x = 204 2] 2 [ y — +]]1y=+x1 = x 2 2

Catalina had her daughter, Amaya, when she was 27 years old. Today, their ages are written with the same digits. Given that Amaya is less than 20 years old, how old are they today? x y – y x = 27 ⇓ (10x + y) – (10y + x) = 27 175

MATHS WORKSH INVESTIGATE LEE E INFÓRMATE

Analyse the following statements and their relationship to the equations and the graph. An amateur cyclist sets off from town A and heads towards town B at a speed of 12 km/h. At the same time, a professional cyclist sets off from B towards A at 24 km/h. At a point between A and B, a retired man is watching the traffic pass by. distance (km) x 0 1/2 5/6 1 30 y = 12x 8

y = 30 – 24x 8

y = 18 + 0x 8

1/2 5/6

1/2

3/2

time (h)

Answer the following questions given that x represents the time since the two cyclists set off and y represents the distance from town A. a) The red line corresponds to the amateur cyclist and the blue line corresponds to the professional. What does the green line represent? b) How long will it take for the cyclists to pass each other? c) How long will it take before the retired man sees the amateur cyclist ride past him? What about the professional? d) What is the distance between A and B? How far from A is the retired man?

READ AND LEARN

Y straight line y=0

An extraordinary system Have you ever wondered which equations are associated with Cartesian axes? Observe that in the equation 0x + 1y = 0, the unknown y takes zero as a value, regardless of the value of x. 0x + 1y = 0 8 y = 0 8

–3

–2

–1

…

It is the equation of the Y axis!

X origin (0, 0)

straight line x=0

• In that case, what is the equation of the Y axis? • And what is the solution of the system formed by these two equations? Other special systems Check that the red and blue lines coincide with the graph for these two equations: What is the solution for the system?

176

x + 0y = 2 x =2 8 * 0x + y = 3 y =3

Remember that you can find academic and professional guidance related to this content at anayaeducacion.es.

Unit 8

ENTRÉNATEMAKES PRACTICE RESOLVIENDO PERFECT!OTROS PROBLEMAS Consider and reflect • All the boys and girls in Guille’s class are going on a field trip. Among other things, they take 14 Spanish omelettes for the trip. At lunch time, every three people share an omelette and during the afternoon snack, each omelette is shared between four people. How many people went on the field trip? • The perimeter of the yellow shape is 160 mm and the area of the blue shape is 600 mm2.

• Find at least three solutions for this sum, taking into account that different letters correspond to different digits.

one one one one + one five

Calculate the area of the yellow shape and the perimeter of the blue shape.

SELF-ASSESSMENT

anayaeducacion.es Answer key and interactive self-assessment.

1 Graph the following equations:

a) y = 2x – 1

7 Write the system shown in the picture and solve it.

b) 2x + 3y – 3 = 0 +

2 Solve the systems with graphs.

a) *

x+ y=7 2x + y = 4 * b) 3x – y = 9 x– y=2

3 Solve by the substitution method.

a) *

x– y=6 x+y= 1 * b) 2x + 3y = 7 3x – y = –9

4 Solve by the equalisation method.

x+y= 2 2x – y = 7 * a) * b) x – y = 10 x+ y=2 5 Solve by the reduction method.

= €8

= €13.75

8 Calculate two numbers if the difference between

them is 119 and three times the smaller number is 17 more than double the bigger number.

9 Yesterday we paid €3 for two coffees and a piece of

toast at a café. However, today we paid €6.30 for three coffees and three pieces of toast. How much is one coffee and how much is a piece of toast?

2x – y = 8 –3x + y = –8 * b) 4x + 5y = 2 x – 2y = 6

10 The base of a rectangle is 8 cm longer than its height

6 Simplify the equations and solve the systems.

11 A shop owner has mixed a high-quality coffee

a) *

Z Z ] x – 3 = y –1 ]] 5x = 2y + 13 ] 3 2 3 a) [ b) [ 2y y 3 ]] 2x = 1 + ] x – =5 5 3 5 4 \ \

Commitment

and the perimeter is 42 cm. What are the dimensions of the rectangle? that costs €7.60/kg, with a lower-quality coffee that costs €4.10/kg, creating a mixture that costs €5.50/kg. How much of each type of coffee has been used?

Watch the video for target 15.c. Think of something you can do to contribute to achieve that goal. Make a commitment to put your idea into practice.

177

PYTHAGOREAN THEOREM Reading and listening

Egyptians and Babylonians The Pythagorean theorem is an important geometrical concept. As you know, it describes the relationship between the squares of the sides of any right-angled triangle. More than 3 000 years ago, both the Egyptians and the Babylonians knew that certain triangles were right-angled and they used them to construct right angles. The Egyptians, for example, used triangles whose sides measured 3, 4 and 5 (which they considered sacred) to divide fields and build pyramids. Pythagoras In his youth, Pythagoras (6th century BCE) travelled to Egypt and Babylon, where he undoubtedly learnt about these properties. His great achievement was to come up with the general theorem describing the relationship between squares drawn on the sides of any right-angled triangle. That is why this theorem bears his name. Euclid's Elements However, it was not Pythagoras who proved the theorem, but Euclid, two centuries later. Euclid of Alexandria wrote Elements around the year 300 BCE. It is a set of 13 books in which he collected, organised and expanded upon all the mathematical knowledge of his time, providing it with a solid logical structure. In Book I he demonstrated what we now know as the Pythagorean theorem. 1 Use context clues to define the phrasal verb ‘come up with’.

Then, write a new sentence in your notebook using this phrasal verb in a different situation. 2 Using past simple and past continuous, write sentences about the

history of the Pythagorean theorem starting with Pythagoras and ending with Euclid.

188

Pythagorean theorem Pythagorean theorem

When the blue triangle is right-angled: A=B+C

The area of the large square is equal to the sum of the areas of the two small ones.

3 What is the area of A if B = 9 cm2 and C = 16 cm2?

Euclid’s proof To prove the Pythagorean theorem, Euclid drew a line perpendicular to the longer side of the triangle (the green line), dividing the large square into two rectangles. He showed that: B = A1 and C = A2. A1 B

A2 C

4 Look at the figure and answer.

Taking a square from the grid as a unit: a) H ow many unit squares does the small square, B, contain? And rectangle A1? What do you notice?

A1 A2

9 B

b) C heck that the number of unit squares in C is the same as in A2.

c) Describe this property.

d) W hat theorem does it confirm? State the theorem.

ANK ANK B E G A U LANG LANGUAGE B ANK ANK GE BANK B B E E G G A A U U GUA K LANG ANG N L A L AN GE BANK 189 ANK GE BANK B B E E G G A A U U G A LAN LANG LANGUA LANGU

5 Draw a right-angled triangle with sides of 5 cm, 12 cm and 13 cm. Write

an equality expressing the relationship between the lengths of the sides.

PYTHAGOREAN THEOREM The two shortest sides of a right-angled triangle form a right angle. They are called the legs. The longest side is called the hypotenuse. In general, we say that a is the hypotenuse and b and c are the legs. According to the Pythagorean theorem: a 2 = b 2 + c 2

This means that the area of the square built on the hypotenuse is equal to the sum of the areas of the squares built on the legs.

This relationship is only true if the triangle is right-angled. a 2 = b 2 + c 2 Interesting fact! This observation was made by the Chinese 400 years before Pythagoras was born. anayaeducacion.es GeoGebra. Graphical demonstration of the Pythagorean theorem.

Look at this demonstration:

The two identical squares have b + c as sides.

If we compare both figures, it is clear that a 2 = b 2 + c 2.

a2 a

a c

Problem solved

What are the areas of the unknown squares in the following shapes?

183 m2

102 m2

47 dm2

S1 386 dm2

As both triangles are right-angled, in both cases the area of the biggest square is equal to the sum of the areas of the smaller squares. Therefore: S1 = 183 m2 + 102 m2 = 285 m2 S2 = 386 dm2 – 47 dm2 = 339 dm2

ãã Pythagorean triples

F ocu s on Eng lish satisfy: to fulfill a condition.

If three natural numbers, c, b, a, satisfy* c 2 + b 2 = a 2, in other words, if they could be the measurements of the sides of a right-angled triangle, we say that the numbers form a Pythagorean triple. Here are a few of them:

3, 4, 5

8, 15, 17

12, 35, 37

5, 12, 13

9, 40, 41

13, 84, 85

7, 24, 25

11, 60, 61

16, 63, 65

Unit 9

ãã The sides of a triangle determine its type Look at the diagrams and analyse the explanation below: Remember

If we know the sides of a triangle, we can find out whether it is right-angled: • If a 2 = b 2 + c 2, the triangle is right-angled. • If a 2 > b 2 + c 2, the triangle is an obtuse triangle. • If a 2 < b 2 + c 2, the triangle is an acute triangle.

right-angled triangle

obtuse triangle

acute triangle

• We already know the relationship between the areas of the squares built on the sides of a right-angled triangle: Figure A 8 52 is equal to 32 + 42 8 In general: a 2 = b 2 + c 2 • If we expand the right angle making it obtuse, the length of the opposite side increases, as does the area of the corresponding square. Therefore: Figure B 8 62 is greater than 32 + 42 8 In general: a 2 > b 2 + c 2 • If we reduce the size of the right angle making it acute, the length of the opposite side decreases, as does the area of the corresponding square. Therefore: Figure C 8 4.52 is less than 32 + 42 8 In general: a 2 < b 2 + c 2 eas

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1 Copy these shapes in your notebook. Draw the square that is missing in each one and say what the area is.

a) b) A1

57 cm2

c) 3 dm2

87 m2

14 dm2

57 cm2

31 m2

A1 = 57 + 57 = … cm2 A2 = 87 + … = … m2 A3 = 14 – … = … dm2 2 Copy and complete to find out whether each of the following triangles is right-angled, acute or obtuse:

a) 70 cm, 240 cm, 245 cm

70 2 + 240 2 = 4 900 + 57 600 = 62 500 4 8 2452 < 702 + 2402 8 The triangle is… 245 2 = 60 025

b) 15 dm, 36 dm, 39 dm

15 2 + 36 2 = 225 + 1296 = f 4 8 392 … 152 + 362 8 The triangle is… 39 2 = 1521

c) 18 m, 80 m, 83 m

18 2 + 80 2 = 324 + 6 400 = 6 724 4 8 832 … 182 + 802 8 The triangle is… 83 2 = f

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CALCULATING ONE SIDE WHEN TWO ARE KNOWN If we know that a triangle is right-angled, and we know the length of two sides, we can use the Pythagorean theorem to calculate the length of the third side.

ãã Calculating the length of the hypotenuse from two known legs Example a?

a 2 = b 2 + c 2 8 a = b 2 + c 2

In a right-angled triangle, the legs are 88 m and 105 m long. Calculate the length of the hypotenuse.

a = 88 2 + 105 2 = 7 744 + 11025 = 18 769 = 137 The hypotenuse is 137 m long.

ãã Calculating one leg when the other leg and the hypotenuse are known Example

b? c

a 2 = b 2 + c 2 8 b 2 = a 2 – c 2 8 b = a 2 – c 2

The hypotenuse of a right-angled triangle is 130 cm long, and one of the legs is 32 cm. Find the length of the other leg.

b = 130 2 – 32 2 = 16 900 – 1024 = 15876 = 126 The other leg is 126 cm long. eas

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1 Copy and complete to find the unknown side in each of these triangles:

The unknown side is the hypotenuse, a. a 2 = 112 + 142 8 a = f 2 + f 2 = f + f = f = … The hypotenuse measures approximately … dm.

11 dm

14 dm

56 km

48 km

The unknown side is the leg, c. 562 = 482 + c 2 8 c 2 = 562 – 482 8 c = f – f = … The other leg measures approximately … km.

Let's practise! 1 Find the length of the unknown side of the following right-angled triangles where

a is the hypotenuse. Round your answer to two decimal places if you need to. a) c = 70 mm; a = 74 mm e) b) b = 15 cm; a = 25 cm 15 cm c) b = 14 m; c = 48 m d) b = 13 inches; c = 84 inches 36 cm

f) 12 cm 37 cm

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ãã Practical exercises with guided solutions Below are various problems which show how the Pythagorean theorem can be used in everyday situations. eas

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Copy and complete these problems. Check that you arrive at the solution given. 2 A zip line is being fitted between two trees that are 12 m apart. The wire will be

attached to one tree at a height of 10 m and to the other at a height of 1 m. How long should the wire be if it needs to be fully taut but with 10 % extra length so that it can be tied to the trees? We know the legs and we have to find the hypotenuse. l 2 = 92 + 122 = … + … = … 8 l = f = … m The length of the taut cable is … m. Adding 10 % 8 … · 1.10 = 16.5 m

10 m

l 1m

12 m

3 We want to use a ramp to get a wheelbarrow up a 1 m step. We have a plank of

d 2 = 2.62 – … = 6.76 – … = … 8 d = f = 2.4 m The foot of the plank should be 2.4 m from the step, or slightly less so that it can rest on top of it.

2.6 d

wood that is 2.6 m long. How far from the step should the ramp start? We know the hypotenuse and the vertical leg. We calculate the other leg.

4 A balloon is tied to the ground with a 20 m rope. It is being blown by the wind,

with the rope pulled taut, so that the balloon is directly above a point on the ground 8 m from where it is tethered. How high is the balloon above the ground? We calculate the length of the vertical leg. h2 = …2 – 82 = … 8 h = f = … m The balloon is … m high.

5 A ladder whose foot is 4 m from the wall reaches a height of 7.5 m. How far from l

8.5 m

7.5 m

the wall would the foot of the ladder have to be for it to reach a height of 8 m? We first need to calculate the length of the ladder. l 2 = 42 + 7.52 = 72.25 8 l = 72.25 = 8.5 m Now we can calculate the distance we are asked to find, d. d 2 = …2 – 82 = … – 64 = … 8 d = f = … m The foot of the ladder should be … m from the wall.

h 2 = 202 – … 8 h = f = … m The height of the attic is … m.

15 5m

h 4m

of his attic. Calculate d , then h. First, we calculate the distance d: d 2 = 52 – 42 = … 8 d = f = … m Now we calculate the height, h, which is the leg of a right-angled triangle. The hypotenuse measures 20 m and the other leg … m:

6 Álvaro has taken the following measurements to find the height, h,

18 m 12 m

anayaeducacion.es GeoGebra. Calculating the unknown side using the Pythagorean theorem. 193

APPLICATIONS  OF THE PYTHAGOREAN THEOREM Many polygons have some elements which are sides of a right-angled triangle. This allows us to associate them using the Pythagorean theorem, and to calculate the length of one of the sides when the other two are known.

eas

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1 The diagonal of a rectangle is 89 cm long, and one of the sides is 80 cm. Calculate

the area. The area of the rectangle with sides a and b is: A = a · b We start by calculating the other side: b = 89 2 – 80 2 = … = … The short side is … cm long. The area is: A = 80 · … = 3 120 cm2

2 The diagonals of a rhombus are 10 cm and 24 cm long. Find the perimeter.

89 cm

80 cm

5 cm

We start by calculating the length of one side:

12 cm

s = 12 2 + … 2 = … = …

Each side is … cm long. The perimeter is: P = 4 · … = 52 cm

3 The side of a rhombus is 6.5 m long and one of the diagonals is 5 m. Find the area.

The area of a rhombus with diagonals d and d' is: A = d · d' 2 We know one of the diagonals. Calculate the other one using the Pythagorean theorem: d' = 6.5 2 – 2.5 2 = … m 2 The second diagonal is … · 2 = … m. Therefore, A = 5 ·… = 30 m2. 2 4 The bases of a right-angled trapezium are 25 cm and  38 cm long, and the height is

19 cm. Find the perimeter. We start by calculating the length of the oblique side: 13 2 + 19 2 =

… ≈… x= The oblique side is approximately … cm long. The perimeter is: P = 38 + 19 + 25 + … = 105 cm

2.5 m

6.5 m

d'/2

25 cm

19 cm 38 cm

x 13 cm

5 Find the area of an isosceles trapezium with bases of 30 cm and 48 cm, and an

oblique side of 41 cm. Remember that the area of a trapezium is: A = (b + b' )· h 2 We start by calculating the height, h. The short side of the green triangle is (48 – 30): 2 = 9 cm. h = 41 2 – 9 2 = … = … The height of the trapezium is … cm. A = (30 + 48) · … = 1 560 cm2 2

b = 30 cm h

41 cm 9 cm

b' = 48 cm

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Unit 9

6 Calculate the area of an equilateral triangle with 8 cm sides.

We begin by calculating the height: h = 82 – 42 = … ≈ … The height is approximately … cm. The area is: A = … · … = 27.6 cm2 2

8 cm

4 cm

7 Calculate the area and perimeter of a regular pentagon with an apothem of 16.2 cm

and a radius of 20 cm. We first calculate the side:

s = … 2 – … 2 = 137.56 ≈ … 2 The side of the pentagon is: s = … · 2 = … cm Therefore, its perimeter is: P = … · 5 = 117 cm Lastly, we calculate the area. Perimeter · apothem … · … = A= = 947.7 cm2 2 2

s/2 16.2 cm

20 cm

8 Find the perimeter of a circumference on which a 6.6 cm chord has been drawn

5.6 cm from the centre. Calculate the area of the corresponding circle. We start by calculating the radius. The shortest side of the coloured right-angled triangle is: k = 6.6 : 2 = 3.3 cm Therefore: r = … 2 + … 2 = 42.25 = … The radius is … cm. P = 2πr = 2 · 3.14 · … ≈ 40.8 cm A = πr 2 = … · …2 ≈ 132.7 cm2

5.6 cm r

6.6 cm

9 A circumference with a radius of  8 cm is intersected by a line at two points, A and

B, which are 8 cm apart. Calculate the area of the circular segment bounded by the % arc AB . The pink circular segment is the difference between the circular sector that intercepts % the arc AB and the triangle OAB. The equilateral triangle OAB is the same triangle whose area we calculated in exercise 6 on this page (Atriangle = 27.6 cm2). 8c m O

m 8c

% Since it is an equilateral triangle, AOB = 60°. Therefore, the area of the sector is one sixth of the area of the whole circle. Asector = (π · …2) : 6 = … cm2 Therefore: Acircular segment = … – … = 5.9 cm2

8 cm

195

3 APPLICATIONS OF THE PYTHAGOREAN THEOREM eas

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10 The distance of a point P to the centre O of a circumference is OP = 23 cm. We

18 cm

r O

23 cm

r 2 = OP 2 – PT 2 = …2 – …2 = … Acircle = πr 2 = … · … ≈ 643.7 cm2 10 cm

11 Calculate the radius of the circumscribed circumference of this square with sides of

10 cm. Remember that the diagonals of the square are perpendicular. Therefore, the coloured triangle is right-angled and has legs of equal length. r 2 + r 2 = …2 8 2r 2 = … 8 r 2 = …

r = f = 7.07 cm D

12 Find the diagonal of a cuboid with dimensions of 1.2 m, 1.6 m and 4.8 m.

Looking at the blue triangle: AC = 1.2 2 + … 2 = … m

1.2 m

Looking at the red triangle: d = … 2 + 4.8 2 = … m The diagonal of the cuboid measures 5.2 m. Note that you can calculate it directly:

B D 4.8 m

d = 1.2 2 + 1.6 2 + 4.8 2 = f = 5.2 m

4.8 m

In general, in a cuboid with dimensions a Ò b Ò c the diagonal is: d = a 2 + b 2 + c 2

1.6 m

1.2 m 1.6 m

13 Calculate the height, h, of a regular pyramid that has a square base with  30 cm

h h

15 m

30 m 30 m

The pyramid is 8 m high. Let’s practise! 1 Find the area of an equilateral triangle with a perimeter

4 A 13-metre chord is drawn on a circumference with

2 Find the area and perimeter of an isosceles trapezium

5 A regular pentagon is inscribed in a circumference

of 54 cm.

with 3.2 m and 6.4 m bases, and 6.3 m height.

3 Calculate the area of a regular hexagon with 18 cm

sides. (Remember that in a regular hexagon the side and the radius have the same length.)

a radius of 9.7 m. How far is the line from its centre?

with a radius of 1 m. Its perimeter is 5.85 m. Calculate the area.

6 Find the length of the diagonal of a cuboid with sides

of 8 dm,  6 dm and 14 dm.

anayaeducacion.es GeoGebra. Calculating areas applying the Pythagorean theorem. 196

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resources

from this un

Unit 9

MS ES AND PROBLE

EXERCIS

Pythagorean theorem 1

Calculate the area of the green square in each of the following cases: a)

60 m2

14 cm2 30 cm2