Section 14.6 solving equations with rational expressions

Slide - 1Copyright © 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G
2
Rational
Expressions and
Applications
14

1. Distinguish between operations with rational
expressions and equations with terms that
are rational expressions.
2. Solve equations with rational expressions.
3. Solve a formula for a specified variable.
Objectives
14.6 Solving Equations with Rational
Expressions

Example
Identify each of the following as an expression or an equation.
Then simplify the expression or solve the equation.
5 7
(a)
9 12
y y
Distinguish Between Expressions and Equations
This is an expression because it
does not have an equals sign.
Simplify by finding a
common denominator
for the two fractions
and adding.
5 7
9 12
y y
20 21
36 36
y y 
41
36
y

24 24
5 5
240
6 8
x x
   
    
   
 
Example (cont)
Identify each of the following as an expression or an equation.
Then simplify the expression or solve the equation.
5 5
(b) 10
6 8
x x 
This is an equation because it has an
equals sign.
Use the multiplication
property of equality.
Multiply by the LCD,
24, and solve.
24 2
5 5
(10)
6 8
4x x
 
  
 

20 15 240x x 
5 240x 
48x 

Uses of the LCD
When adding or subtracting rational expressions,
keep the LCD throughout the simplification.
When solving an equation with terms that are
rational expressions, multiply each side by the LCD
so that the denominators are eliminated.

Example Solve. Check the proposed solution.
2
4
4 4
y y
y y

 
Solve Equations with Rational Expressions
Multiply both sides by (y – 4).
2
4
4) 4
4
( ( )
4
y y
y y
y y
   
   




2
4y y
2
4 0y y 
( 4) 0y y  
y = 0 or y – 4 = 0
y = 4
Checking: y = 4 would
cause the denominator in the
original problem to equal 0.
Therefore, the only solution
to this equation is y = 0.

Solve Equations With Rational Expressions
Solving an Equation with Rational Expressions
Step 1 Multiply each side of the equation by the
LCD. (This clears the equation of fractions).
Be sure to distribute to every term on both
sides of the equation.
Step 2 Solve the resulting equation for proposed
solutions.
Step 3 Check each proposed solution by
substituting it in the original equation. Reject
any that cause a denominator to equal 0.

2
1 2
9 3 3
x
x x x
 
  
Factor the denominators.
  
1 2
3 3 3 3
x
x x x x
 
   
Multiply each side of the equation by the LCD, (x + 3)(x – 3).
  
1 2
3 3 3
( 3)( 3) ( 3)( 3)
3
x
x
x
x
x x x
x x
 
  
  
  
 


Example (cont) Solve. Check the proposed solution.
x + x – 3 = 2x + 6
2x – 3 = 2x + 6
–3 = 6
Checking: The statement –3 = 6 is
false.
x + (x – 3) = 2(x + 3)
  
1
3 3
( 3)( 3)
3
( 3)( 3)x
x
x x
x
x x
x   


 

( 3 )
3
)( 3
2
x x
x



The solution set is 0.

2
1 1 2
3 10 2 4 2 10z z z z
 
   
Factor the denominators.
1 1 2
( 5)( 2) 2( 2) 2( 5)z z z z
 
   
Multiply each side of the equation by the LCD, 2(z + 2)(z – 5).
1 1 2
( 5)( 2) 2( 2) 2( 5)
2( 2)( 5) 2( 2)( 5)z
z
z
z
z
z
z
z
   
         
  



Example (cont) Solve. Check the proposed solution.
2 + z – 5 = 2z + 4
z – 3 = 2z + 4
–3 = z + 4
–7 = z
Checking: z = –7 does not make
any denominator equal 0.
The solution set is {–7}.
2 + (z – 5) = 2(z + 2)
1 1
( 5)
2(
( 2)
2)( 5
2( 2)
) 2( 2)( 5)
z z
z z z z
z
   
      


 


2
2
2
(
( 2)(
5)
5)z
z
z
 
  




Example Solve the formula for the specified variable.
1 1 1
for .
2
y
x y x
 
Solve a Formula for a Specified Variable
Multiply by the LCD, 2xy.
1
2
1
2
1
2x y
xy y
x
x
 
   
 

2y – 2x = y
2y – y = 2x
y = 2x
Distributive property
1 1 1
2
2
2 2xy xy x
x y
y
x
 
  
    


  
Simplify.
Subtract y and add 2x.
Solve.

Section 14.6 solving equations with rational expressions

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