# 11-5

```11-5
Solving Rational Equations
Algebra I
Hint for solving:
• A quick way to solve a rational
equation is to multiply everything
through by the LCD.
• This will get rid of all the fractions!
• Beware of extraneous solutions!
Example 1: Solve.
3 1 12
 
x 2 x
(2 x)3 (2 x) (2 x)12


x
2
x
6  x  24
 x  18
x  18
LCD: 2x
Multiply each fraction
through by the LCD
3
1 12
 
 18 2  18
3  9  12
Ex. 2
Solve.
5x
5
 4
x 1
x 1
LCD: ?
LCD: (x+1)
5 x( x  1)
5( x  1)
 4( x  1) 
( x  1)
( x  1)
5(1)
5
 4
11
11
5x  4 x  4  5
5x  4 x  1
x  1
5
5
 4
0
0
No Solution!
?
Ex. 3: Solve.
3
1

2
x  4x x  4
x  4 x  3( x  4)
2
x  4 x  3 x  12
2
x  x  12  0
2
( x  4)( x  3)  0
x  4  0 or x  3  0
x  4 or x  3
Assignment
Ex. 3 3x  2  6  1
2
x

2
x
4
Solve.
Factor 1st!
3x  2
6

1
x  2 ( x  2)( x  2)
LCD: (x+2)(x-2)
(3x  2)( x  2)( x  2) 6( x  2)( x  2)

 ( x  2)( x  2)
( x  2)
( x  2)( x  2)
3x 2  6 x  2 x  4  6  x 2  2 x  2 x  4
3x 2  4 x  4  x 2  2
2x  4x  6  0
2
x  2x  3  0
( x  3)( x  1)  0
x  3 or x  1
2
x  3  0 or x 1  0
Short Cut!
When there is only one fraction on each
side of the =, just cross multiply as if
you are solving a proportion.
Ex. 5: Solve.
6
x2

2
2x  2x x 1
6
x2

2 x( x  1) x  1
2 x( x  1)( x  2)  6( x  1)
2 x( x  2)  6
x( x  2)  3
2
x  2x  3  0
( x  3)( x  1)  0
x  3  0 or x 1  0
x  3 or x  1