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Chapter 6 Section 6
Solving Rational
Equations
Solving Rational Equations
with Integer in Denominator
A rational equation is one that contains one or
more rational (fractional) expressions.
Rational Coefficient
1
3
x x 8
2
5
Rational Term
variable in denominator
3
5
x2
x 3x

8
2 5
Solving Rational Equations
1. Determine the LCD of all fractions
2. Multiply both sides (every term) of the equation by the LCD.
Eliminates the fraction
3. Remove any parentheses and combine like terms on each
sides
4. Solve using the properties learned earlier, denominator ≠ 0
5. Check your solution with original equation
Solving Rational Equations
with Integer in Denominator
The strategy used to solve rational equations is to change the
rational equation into the kind of equation we can solve.
Example:
x x
 1
2 3
LCD = (2)(3) = 6
x x
(6)     (6)(1)
2 3
3

 x
 x
(6)    (6)    (6)(1)
2
3
2
6x 6x

6
2
3

3x  2 x  6

x6
Solving Rational Equations
with Integer in Denominator
Example: Solve
x2 1 x2
 
10 2 20
LCD = 20
 x2
1 x2
(20) 
  (20)  

10
2
20




(2)( x  2)  (10)(1)  ( x  2)
x  12
 x  2  10  1 
 x2
( 20 ) 
  ( 20 )    ( 20 ) 

2
 10 
 20 
2


2x  4  10  x  2
Distributive Property

2 x  x  10  2  4
Combine like terms.
Variable on one side.
Solving Rational Equations with a
variable in the denominator
 Remember that dividing by zero is
undefined. Therefore, you must check your
answer in the original equation. If it makes
the denominator zero it is not a solution.
 This is called an extraneous roots or
extraneous solution.
 It is only necessary to check the
denominator.
Solving Rational Equations with a
variable in the denominator
Example: Solve
5 7
4 
x 3
LCD = 3x
5

7
(3x)  4    (3 x)  
x

3
12 x  15  7 x
x3


5
7
(3 x)(4)  (3 x )    ( 3 x)  
 x
3
12x  7 x  15

5x  15
Solving Rational Equations with a
variable in the denominator
Example: Solve
m2 2

m 1 3
LCD = (m+1)(3)
Distributive Property
 m2 
2
(3)( m  1 ) 
  ( 3 )(m  1)  
 m 1 
3
3m  6  2m  2


3m  2m  2  6
(3)(m  2)  (m  1)  2 

m8
Solving Rational Equations with a
variable in the denominator
Remember that with Proportions we can cross multiply.
a c

b d

Example: Solve
a
b

c
d
3
2

z2 z4

use cross multiplication
(3)(z  4)  (2)( z  2)
3z  2 z  4  12
ad  bc


3 z  12  2 z  4
z  16
Solving Rational Equations with a
variable in the denominator
Example: Solve
5
x 4
x
LCD  x
5

( x)  x    ( x)(4)
x

x2  5  4 x
x 5  0
x5

and

 5 
( x)( x)  ( x )    4 x
 x 
x2  4 x  5  0
x 1  0
x  1

(x  5)(x  1)  0
Solving Rational Equations with a
variable in the denominator
Example: Solve
x2  5x
14

x2
x2
LCD  x  2
 x2  5x 
 14 
( x  2 )

(
x

2
)



x

2
x

2




x  5 x  14  0

x7  0
x20
2
x7
and

(x  7)(x  2)  0
x  2
( x 2  5 x)  14
-2 makes the
denominator zero,
therefore, -2 is an
extraneous and is NOT
a solution
Solving Rational Equations with a
variable in the denominator
Example: Solve
7x
3
5


x2  9 x  3 x  3
Factor: x 2  9  ( x  3)( x  3)
LCD  (x  3)(x  3)


 3 
 5 
7x
( x  3 )( x  3 ) 
  ( x  3)( x  3 ) 
  ( x  3 )( x  3) 

(
x

3
)(
x

3
)
x

3
x

3






7 x  ( x  3)(3)  ( x  3)(5)
4 x  9  5 x  15


7 x  (3x  9)  5 x  15
- 9  15  5 x  4 x

6x

7x  3x  9  5 x  15
Remember
 When adding and subtracting rational
expressions, find the LCD and rewrite each
expression with the common denominator.
 When solving a rational equation, multiply by the
LCD to eliminate the fractions.
 Make sure that the equation is in standard form.
ax2 + bx + c = 0
 Always determine the values for the variables
that will make the denominator equal zero. This
is so that you can spot an extraneous solution.
HOMEWORK 6.6
Page 394-395:
#15, 25, 33, 39, 41, 49, 53, 61