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Solution and Answer Guide: Larson, College Algebra, 12e, 2027, 9798214407401; Chapter 1: Equations, Inequalities, and Mathematical Modeling

Solutions Manual for College Algebra 12th Edition By Ron Larson

Solution and Answer Guide LARSON, COLLEGE ALGEBRA, 12E, 2027, 9798214407401; CHAPTER 1: EQUATIONS, INEQUALITIES, AND MATHEMATICAL MODELING

TABLE OF CONTENTS Section 1.1 Graphs of Equations ........................................................................................ 95 Section 1.2 Linear Equations in One Variable ................................................................. 109 Section 1.3 Modeling with Linear Equations .................................................................. 129 Section 1.4 Quadratic Equations and Applications ........................................................ 145 Section 1.5 Complex Numbers .......................................................................................... 181 Section 1.6 Other Types of Equations ............................................................................. 192 Section 1.7 Linear Inequalities in One Variable .............................................................. 225 Section 1.8 Other Types of Inequalities .......................................................................... 239 Review Exercises for Chapter 1 ....................................................................................... 272 Problem Solving for Chapter 1 ......................................................................................... 303

Solutions Manual for College Algebra 12th Edition By Ron Larson

© 2027 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

94


Solution and Answer Guide: Larson, College Algebra, 12e, 2027, 9798214407401; Chapter 1: Equations, Inequalities, and Mathematical Modeling

Solutions Manual for College Algebra 12th Edition By Ron Larson

SECTION 1.1 GRAPHS OF EQUATIONS 1.1.1. solution or solution point 1.1.2. graph 1.1.3.

Three approaches to solve problems mathematically are algebraic, graphical, and numerical. 1.1.4.

( ) and ( x, y ) is the radius r. So, ( x − h) + ( y − k ) = r .

(

Let x, y be any point on the circle. The distance between the center h, k 2

)

2

1.1.5.

(

) ()

2

(a) 2, 0 : 2

()

?

− 3 2 + 2=0 ?

4 − 6 + 2=0 0=0 Yes, the point is on the graph.

(

?

) ( ) − 3 ( −2) + 2 = 8

(b) −2, 8 : −2

2

?

4+6+2 =8 12  8 No, the point is not on the graph. 1.1.6.

(

)

?

(a) 0, 2 : 2 = 0 + 4

2=2 Yes, the point is on the graph.

(

)

?

(b) 5, 3 : 3 =

5+4

?

3= 9 3 = 3 Solutions Manual for College Algebra 12th Edition By Ron Larson Yes, the point is on the graph.

© 2027 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

95


Solution and Answer Guide: Larson, College Algebra, 12e, 2027, 9798214407401; Chapter 1: Equations, Inequalities, and Mathematical Modeling

Solutions Manual for College Algebra 12th Edition By Ron Larson

1.1.7. ?

( )

(a) 1, 5 : 5 = 4 − 1 − 2 ?

5 = 4−1 53 No, the point is not on the graph.

(

?

)

(b) 6, 0 : 0 = 4 − 6 − 2 ?

0 = 4−4 0=0 Yes, the point is on the graph. 1.1.8.

(

2 ?

) ( ) + 5 (0 ) = 8 2 ( 36) + 5 (0) = 8

(a) 6, 0 : 2 6

2

?

?

72 + 0 = 8 72  8 No, the point is not on the graph.

(

2 ?

) ( ) ( ) 2 (0) + 5 ( 16) = 8

(b) 0, 4 : 2 0

2

+5 4 = 8 ?

?

0 + 80 = 8 80  8 No, the point is not on the graph. 1.1.9.

y = −2x + 5 x

−1

0

1

2

5 2

y

7

5

3

1

0

( x, y ) ( −1, 7) (0, 5) ( 1, 3) (2, 1)

5   , 0 2 

Solutions Manual for College Algebra 12th Edition By Ron Larson

© 2027 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

96


Solution and Answer Guide: Larson, College Algebra, 12e, 2027, 9798214407401; Chapter 1: Equations, Inequalities, and Mathematical Modeling

Solutions Manual for College Algebra 12th Edition By Ron Larson

1.1.10.

y=

3 x−1 4

x

−2

0

1

y

−

5 2

–1

−

( x, y )

 5  −2, −  2 

(0, − 1)

 1  1, −  4 

1 4

4 3

2

0

1 2

4   , 0 3 

 1  2,   2

1.1.11.

y = x 2 − 3x x

−1

0

1

2

3

y

4

0

–2

–2

0

( x, y )

( −1, 4)

(0, 0)

( 1, − 2)

(2, − 2)

(3, 0)

Solutions Manual for College Algebra 12th Edition By Ron Larson

© 2027 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

97


Solution and Answer Guide: Larson, College Algebra, 12e, 2027, 9798214407401; Chapter 1: Equations, Inequalities, and Mathematical Modeling

Solutions Manual for College Algebra 12th Edition By Ron Larson

1.1.12.

y = 5 − x2 x

−2

–1

0

1

2

y

1

4

5

4

1

( −1, 4)

(0, 5)

( 1, 4)

(2, 1)

( x, y ) ( −2, 1)

1.1.13.

( ) y-intercept: (0, 2) x-intercept: −2, 0

1.1.14.

(

x-intercept: 4, 0

(

)

)

y-intercepts: 0,  2 1.1.15.

( ) Solutions Manual for College Algebra 12th Edition By Ron Larson y-intercept: (0, 9) x-intercept: 3, 0

© 2027 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

98


Solution and Answer Guide: Larson, College Algebra, 12e, 2027, 9798214407401; Chapter 1: Equations, Inequalities, and Mathematical Modeling

Solutions Manual for College Algebra 12th Edition By Ron Larson

1.1.16.

( )( ) y-intercepts: (0, 16) , (0, 0) x-intercepts: 2, 0 , 0, 0

1.1.17.

x2 − y = 0

( − x ) − y = 0  x − y = 0  y-axis symmetry x − ( − y ) = 0  x + y = 0  No x-axis symmetry ( − x ) − ( − y ) = 0  x + y = 0  No origin symmetry 2

2

2

2

2

2

1.1.18.

x − y2 = 0

( − x ) − y = 0  − x − y = 0  No y-axis symmetry x − ( − y ) = 0  x − y = 0  x-axis symmetry ( − x ) − ( − y ) = 0  − x − y = 0  No origin symmetry 2

2

2

2

2

2

1.1.19.

y = x3

( )  y = − x  No y-axis symmetry

y = −x

3

3

− y = x 3  y = − x 3  No x-axis symmetry

( )  − y = − x  y = x  Origin symmetry

− y = −x

3

3

3

1.1.20.

y = x4 − x2 + 3

( ) − ( −x ) + 3  y = x − x + 3  y-axis symmetry

y = −x

4

2

4

2

− y = x 4 − x 2 + 3  y = − x 4 + x 2 − 3  No x-axis symmetry

( ) − ( − x ) + 3  y = − x + x − 3  No origin symmetry

− y = −x

4

2

4

2

Solutions Manual for College Algebra 12th Edition By Ron Larson

© 2027 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

99


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