Intermediate Algebra (11th edition)

SECTION 7.4 Equations with Rational Expressions and Graphs 391

Complete solution available
on the Video Resources on DVD

7.4 EXERCISES

As explained in this section, any values that would cause a denominator to equal 0 must be

excluded from the domain and, consequently, as solutions of an equation that has variable

expressions in the denominators.

(a)Without actually solving each equation, list all possible values that would have to be

rejected if they appeared as proposed solutions.

(b)Then give the domain, using set-builder notation.

See Example 1.

1. 
   
   
   
   2. 
      
   
   
   
   

3. 4.

5. 6.

7. 8.

9. 10.

11. 12.

13.Suppose that in solving the equation

all of your algebraic steps are correct. Is there a possibility that your proposed solution

will have to be rejected? Explain.

14.Consider the equation in Exercise 13.

(a)Solve it. (b)Check your solution, showing all steps.

x+ 7

4

-

x+ 3

3

=

x

12

,

4 x- 1

2 x+ 3

=

12 x- 25

6 x- 2

3 x+ 1

x- 4

=

6 x+ 5

2 x- 7

4

3 x- 5

+

2

x

=

9

4 x+ 13

6

4 x+ 7

-

3

x

=

5

6 x- 13

3

x^2 +x

-

1

x+ 5

=

2

x- 7

2

x^2 - x

+

1

x+ 3

=

4

x- 2

2

x^2 - 25

-

1

x+ 5

=

1

x- 5

1

x^2 - 16

-

2

x- 4

=

1

x+ 4

3

x+ 4

-

2

x- 9

= 0

1

x+ 1

-

1

x- 2

= 0

5

6 x

-

8

2 x

=

x

4

1

3 x

+

1

2 x

=

x

3

For Discussion or Writing

Use each graph to determine the solution set of the equation.

1. 2.

3

–3

–3 3

5

–2

–3 3

ƒ 1 x 2 =

2

x

-

1

x^2

ƒ 1 x 2 =

1

x^3

+

1

x

+ 2

ƒ 1 x 2 = 0