This document discusses how to solve rational equations. It defines a rational equation as an equation containing one or more rational expressions. It provides the steps to solve rational equations which are: 1) find the least common multiple (LCM) of the denominators, 2) multiply every term by the LCM, 3) cross cancel, and 4) solve the resulting linear or quadratic equation. It notes that solving a rational equation may result in an extraneous solution so all solutions must be checked in the original equation. Several examples of solving rational equations are then worked through.

1. 10.6 Solving Rational Equations

2. Rational Equation
An equation containing one or more rational
expressions
2
2 1 5
3
x x
x x
 
 

3. Steps to solve Rational Equations
1. Find the LCD
2. Multiply every term on both sides of the
equation by the LCD over 1
3. Cross Cancel
4. Solve for the variable
a. Linear equation: get variables on one side and
constants on the other
b. Quadratic: set your equation = 0 and factor.

4. Extraneous Solutions
• When both sides of the equation are mult by
a variable, the equation is transformed into
a new equation and may have an extra
solution.
• Check each solution in the original rational
equation
• Make sure that your answer does not make
the denominator 0

5. Solving Rational Equations
Multiply both sides of the equation
by the LCM of the denominators.
x
x 

4
1
1
Least Common Multiple: Each factor
raised to the greatest exponent.
 
 x
x 4
 
LCM is 4
x x

 
x
x 
 4
x
x 

4
x
2
4  x

2

6. Solve for x:
12
8
5
4
3 x

 LCM =
x
2
15
18 

x

2
33
2
2 3
2
3
2 3 24
 
24
1
24
1

2
2 3


7. Solve for x:
x
x
2
1
1


LCM =

x
(x + 1)(x)
(x + 1)(x)• •(x + 1)(x)
2
2 
x
x

2
1x
 1x

0 2
x
 

8. Solve for x:
12
1
2
1



x
x
LCM =

 2
1
2x
2x• •2x
x
24

x


8
1

9. Solve for x:
2
1


x
x

1
2
x
x• • x
x
2
1

x
0
1
2
2


 x
x
  0
1
2


x

10. Solve for x:
2
4
2
2


 x
x
x
2

x
4
2

x
2


x
-2 is an extraneous solution.

11. Solve for x:
2
4
2
2


 x
x
x
2

x
4
2

x
2


x
-2 is an extraneous solution.

12. Cross products
Short cut:
7 1
2 4
x
x



 
4 7
x  
extraneous solution?
 
1 2
x 
4 28 2
x x
  
1x
 1x

3 28 2
x 
3 30
x 
10
x 

13. Cross products: 2 1
1 2
x x

 
1 2 4
x x
  
5 x


14. Cross products: 3 1
3 2 5
x
x



5 15 3 2
x x
  
2 17
x 
17
2
x 

15. Cross products: 2 1
3 1
x x
x x
 

 
2
2
x x
   2
4 3
x x
 
2
x
 2
x

2 4 3
x x
    
3 2 3
x  5
3
x 
4x
 4x

3 5
x 

16. Assignment:
Page 453
# (2-24) even