8.6 Solve Rational Equations

Goal  Solve rational equations.

VOCABULARY
Cross multiplying
A method of solving a simple rational equation for which each side of the equation is a single rational expression. Equal products are formed by multiplying the numerator of each expression by the denominator of the other.

Example 1

2

5

5

2

20

-

=

-

x

x

Solve a rational equation by cross multiplying
	Original equation	
20 __(x  2)__ = 5 __(3x  5)__	Cross multiply.	
__20 x  40__ = __15x  25__	Distributive property	
__5x  40__ = __25__	Subtract __15 x__ from each side.	
__5 x__ = __15__	Add __40__ to each side	
__x__ = __3__	Divide each side by __5__.	

-

=

-

5

)

3

(

3

20

5

3

20

x

CHECK
 = __5__ and

2

3

5

2

5

-

=

-

x

 = __5__

 Checkpoint Solve the equation by cross multiplying.

8

3

6

2

3

+

=

+

x

x

1.	
 4

1

12

2

6

-

-

=

+

-

x

x

2.	
 5

Your Notes

Example 2
Solve a rational equation with one solution

x

x

14

3

11

8

-

=

+

÷

ø

ö

ç

è

æ

-

x

14

÷

ø

ö

ç

è

æ

+

3

11

8

x

	Original equation	
__3x__ 	.= __3x__ 		Multiply each side by the LCD, __3x__.	
__24 +11x__ = __ 42__	Simplify.	
__x__ = __6__	Solve for __x__ .	
The solution is __6__. Check this in the original equation.

Example 3
Solve a rational equation with two solutions

)

4

)(

3

(

8

2

3

5

3

-

+

+

=

+

+

x

x

x

x

Solve: 	

)

4

)(

3

(

8

2

3

5

3

-

+

+

=

+

+

x

x

x

x

The LCD is __(x + 3)(x  4)__ .

÷

ø

ö

ç

è

æ

+

+

3

5

3

x

x

÷

÷

ø

ö

ç

ç

è

æ

-

+

+

)

4

)(

3

(

8

2

x

x

__(x + 3)(x  4)__ 	 = __(x + 3)(x  4)__ 	
__(x  4)(3x  5)__ = __2(x + 3)(x  4) + 8__
__3x2  17x + 20__ = __2x2  2x  16__
__x2  15x + 36__ = 0
__(x  3)(x  12)__ = 0
x = __3__ or x = __12__
The solutions are __3__ and __12__ . Check these in the original equation.

 Checkpoint Solve the equation by using the LCD.

x

 

 

x

 

 

2

5

3

4

11

=

-

3.	
2

4.	
x

 

 

 

x 

 

 

9

2

4

1

=

+

+

3, 6

Your Notes

Example 4
Check for extraneous solutions

4

4

4

3

16

6

2

2

 

 

x

 

 

x

x

 

 

x

x

-

=

+

-

-

Solve:	
The LCD is __(x + 4)(x  4)__.

4

4

-

x

ú

û

ù

ê

ë

é

+

-

-

+

4

3

)

4

)(

4

(

6

2

x

x

x

x

x

4

4

4

3

)

4

)(

4

(

6

2

-

=

+

-

-

+

x

x

x

x

x

x

__(x + 4)(x  4)__ 	 = __(x + 4)(x  4)__  	
__6x2 3x2 + 12x__ = __4x + 16__
__3x2 + 8x  16__ = 0
__(3x  4)(x + 4)__ = 0

3

4

__3x  4__ = 0 or __x + 4__ = 0
x = ____ or x = __ 4__

3

4

You can use algebra to check whether either solution is extraneous.
CHECK The solution ____ checks, but the apparent solution __ 4__ is extraneous, because substituting it in the equation results in division by zero , which is undefined .

 Checkpoint Solve the equation by using the LCD. Check for extraneous solutions.

3

2

3

4

9

8

2

2

-

=

+

-

-

x

 

x 

x

 

 

x

x

5.	

2

1

Homework
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