Section 9-6

Solving Rational Equations and Inequalities

4

3

2

3

28

9

z

12

11

4

12

d

2

1

4

1

3

1vv

5

2

2

pp

Ryann Noe

• Rational equation:

65500

x

x

• Rational Inequalities:

2

1

8

5

4

1

aa

any equation that contains one or more rational expressions

Ex:

Inequalities that contain one or more rational expressions

Ex:

Solving Rational Equations

• Easier to solve once fractions are eliminated

• Eliminate by using Least Common Denominator

FRACTIONS

x6 x66

1

3

2

2

11xx

Multiply out

33-4=x

x=29 Answer!

Find LCD then Solve 4

3

2

3

28

9

z Check your solution

LCD= 28(z+2)

4

3

2

3

28

9

z

multiply

(9z+18) + 84 = 21z + 42

9z + 102 = 21z + 42

60 = 12z

5 = z

CHECK

4

3

2

3

28

9

z

4

3

25

3

28

9

4

3

7

3

28

9

4

3

28

12

28

9

4

3

4

3

True so:

Z=5

Solving Inequalities

Excluded values:

1. State the excluded values

2. Change the sign to equals and solve

3. Use the values found by solving to make a number line and then test points

values that make the denominator equal to zero

example:2

1

8

5

4

1

aa

1. The only excluded value is 0

2. Solve:

2

1

8

5

4

1

aa

2+5=4a7=4aa= 4

31

a8 a8

On your number line draw vertical lines at the excluded number and your solution to separate into regions

-3 -2 -1 0 1 2 3

Then test a value in each section and see which number comes out true

Test a= –1

2

1

)1(8

5

)1(4

1

2

1

8

5

4

1

2

1

8

7

This is NOT true so a<0 is not a solution

Test a=1

2

1

)1(8

5

)1(4

1

2

1

8

5

4

1

2

1

8

7

True so: 0<a< 4

31

Test a=2

2

1

)2(8

5

)2(4

1

2

1

16

5

8

1

2

1

16

7

This is NOT true so a> is not a solution4

31

This means that the final answer is

0<a<4

31

Homework

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