The document outlines the steps for solving rational equations, emphasizing the importance of finding the least common denominator (LCD) and addressing potential extraneous solutions. It details the processes of multiplying each term by the LCD, cross-canceling, and solving for variables using various methods, including factoring for quadratic equations. Additionally, it includes specific examples and reminders to check each solution against the original equation to avoid invalid results.

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
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:
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