Transcript Solve equations with rational expressions.

6.6 Solving
Equations with
Rational
Expressions
Objective 1
Distinguish between operations with
rational expressions and equations
with terms that are rational
expressions.
Slide 6.6-3
Distinguish between operations with rational expressions
and equations with terms that are rational expressions.
Before solving equations with rational expressions, you must understand the
difference between sums and differences of terms with rational coefficients,
or rational expressions, and equations with terms that are rational
expressions.
Sums and differences are expressions to simplify. Equations are solved.
Uses of the LCD
When adding or subtracting rational expressions, keep the LCD throughout
the simplification.
When solving an equation, multiply each side by the LCD so the denominators
are eliminated.
Slide 6.6-4
CLASSROOM
EXAMPLE 1
Distinguishing between Expressions and Equations
Identify each of the following as an expression or an equation. Then
simplify the expression or solve the equation.
x x 5
 
2 3 6
Solution:
equation
 x x 5
6     6
2 3 6
3x  2 x  5
x 5
5
2x 4x

3
9
expression
3 2x 4x
  
3 3
9
6x 4x


9
9
2x

9
Slide 6.6-5
Objective 2
Solve equations with rational
expressions.
Slide 6.6-6
Solve equations with rational expressions.
When an equation involves fractions, we use the multiplication property of
equality to clear the fractions. Choose as multiplier the LCD of all
denominators in the fractions of the equation.
Recall from Section 6.1 that the denominator of a rational expression cannot
equal 0, since division by 0 is undefined. Therefore, when solving an
equation with rational expressions that have variables in the denominator,
the solution cannot be a number that makes the denominator equal 0.
Slide 6.6-7
CLASSROOM
EXAMPLE 2
Solving an Equation with Rational Expressions
Solve, and check the solution.
2m  3 m
6
 
5
3
5
Check:
2m  3 m
6
 
5
3
5
 2m  3 m   6 
 2  9   3  9  
15 
     15
6

3   5  15  
   15
 5
5
3 
5

6m  9  5m  18
3  21  5  3  18
Solution:
m 9  9  18  9
m  9
63  45  18
18  18
The use of the LCD here is different from its use in Section 6.5. Here, we use the
multiplication property of equality to multiply each side of an equation by the LCD.
Earlier, we used the fundamental property to multiply a fraction by another fraction
that had the LCD as both its numerator and denominator.
Slide 6.6-8
Solve equations with rational expressions. (cont’d)
While it is always a good idea to check solutions to guard against arithmetic
and algebraic errors, it is essential to check proposed solutions when
variables appear in denominators in the original equation.
Solving an Equation with Rational Expressions
Step 1: Multiply each side of the equation by the LCD to clear the equation
of fractions. Be sure to distribute to every term on both sides.
Step 2: Solve the resulting equation.
Step 3: Check each proposed solution by substituting it into the original
equation. Reject any that cause a denominator to equal 0.
Slide 6.6-9
CLASSROOM
EXAMPLE 3
Solving an Equation with Rational Expressions
Solve, and check the proposed solution.
2
2x
1

x 1 x 1
Solution:
2  2x

 x  1 1 
 x  1

 x 1  x 1
 x  1  2  x  2x  x
1  x
When the equation is solved, − 1 is a proposed solution. However, since x =
− 1 leads to a 0 denominator in the original equation, the solution set is Ø.
Slide 6.6-10
CLASSROOM
EXAMPLE 4
Solving an Equation with Rational Expressions
Solve, and check the proposed solution.
2
3
 2
2
p 2p p  p
Solution:

 

2
3
p  p  2  p  1 
  
 p  p  2  p  1
 p  p  2    p  p  1 
2 p  2  2p  3p  6  2p
2  6  p  6  6
p4
The solution set is {4}.
Slide 6.6-11
CLASSROOM
EXAMPLE 5
Solving an Equation with Rational Expressions
Solve, and check the proposed solution.
8r
3
3


2
4r  1 2r  1 2 r  1
Solution:

  3
8r
3 

 2r  1 2r  1 
  
  2r  1 2r  1
  2r  1 2r  1   2r  1 2r  1 
8r  6r  3  6r  3
8r 12r  12r 12r
4r  0
r 0
Since 0 does not make any denominators equal 0, the solution
set is {0}.
Slide 6.6-12
CLASSROOM
EXAMPLE 6
Solving an Equation with Rational Expressions
Solve, and check the proposed solution (s).
1
1
2
 
x  2 5 5  x2  4
Solution:
 1

1 
2
5  x  2  x  2  
   
 5  x  2  x  2 
  x  2  5   5  x  2  x  2  
5 x  10  x 2  4  2  2  2
x2  5x  4  0
 x  1 x  4  0
x  1 1  0 1
x  1
or
x44  04
x  4
The solution set is {−4, −1}.
Slide 6.6-13
CLASSROOM
EXAMPLE 7
Solving an Equation with Rational Expressions
Solve, and check the proposed solution.
6
1
4

 2
5 x  10 x  5 x  3x  10
Solution:


6
1  
4
5  x  2  x  5  

  
 5  x  2  x  5 
 5  x  2   x  5     x  2  x  5  
6 x  30   5x  10  20
x  40  40  20  40
x  60
The solution set is {60}.
Slide 6.6-14
Objective 3
Solve a formula for a specified variable.
Slide 6.6-15
CLASSROOM
EXAMPLE 8
Solving for a Specified Variable
Solve each formula for the specified variable.
st
b
for s
r
Solution:
st  r 
b( r ) 
 
r 1
br  s  t
b t  x t t
br  t  s
x
z
for y
x y
x
z  x  y 
 x  y
x y
z  x  y x

z
z
x yx 
x
x
z
x
y x
z
Remember to treat the variable for which you are solving as if it were the only
variable, and all others as if they were contants.
Slide 6.6-16
CLASSROOM
EXAMPLE 9
Solving for a Specified Variable
Solve the following formula for z.
2 1 1
 
x y z
Solution:
2  1 1
xyz       xyz
 x  y z
2 yz xz  xy

z
z
xy
2y  x  x   x
z
 1 
 xy   1 
   2 y  x     
 z   xy 
 xy 
1 2y  x

z
xy
xy
z
2y  x
When solving an equation for a specified variable, be sure that the specified variable
appears alone on only one side of the equals symbol in the final equation.
Slide 6.6-17