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M098
Carson Elementary and Intermediate Algebra 3e
Section 7.6
Objectives
1.
Solve equations containing rational expressions.
Vocabulary
Extraneous
solution
An apparent solution that does not solve its equation.
Prior Knowledge
1. Difference between equations and expressions.
It is important to remember the difference between expressions and equations. Expressions can be
simplified but they can not be solved. Equations, on the other hand, can be solved.
2. Solve linear equations with fractions.
7x
x5
1


12
8
4
 7x 
 x  5
 1
24   24
  24 
12
8
 


4
Clear the fractions by multiplying each term by the LCD,
which is 24.
2(7x) – 3(x – 5) = 6(1)
Reduce each term.
14x – 3x + 15 = 6
11x + 15 = 6
Remove parentheses by multiplying and using the
distributive property. Be careful to distribute the minus
to BOTH terms in the parentheses.
Combine like terms
11x = -9
Subtract 15 from both sides.
x  
9
11
Divide both sides by 11.
New Concepts
1. Solve equations containing rational expressions.
We spent quite a little time in Chapter 2 working with equations that contained fractions. Solving a
rational equation with a variable in the denominator is solved the same way with the addition of one
step. We must keep in mind that the denominator can not be equal to zero. As soon as you determine
the LCD, indicate the restrictions that must be placed on your results; those numbers that will cause
the denominator to equal zero. If one of these numbers should be one of the solutions when you solve
the equation, it must be eliminated. These are referred to as extraneous solutions.
Example 1: What are the restrictions for
3
2
  7x ?
x5 3
x can’t equal 5 because 5 causes the denominator of the first fraction to be 0, which
makes the fraction undefined.
V. Zabrocki 2011
page 1
M098
Carson Elementary and Intermediate Algebra 3e
Example 2: What are the restrictions for
Section 7.6
3x 2
  7x ?
5
3
There are no restrictions because there are no variables in the denominator. The
fractions will never be undefined.
5y
Example 3: Solve
5y
y  3y  1
5y
y  3y  1

2
y  2y  3
2
y  2y  1
y  3y  1y  2 

2

2
2
y y2
3y
y  2y  3
y  2y  1
3y
2
y  5y  6
.
Factor each denominator to find the LCD.
LCD = (y + 3)(y – 1)(y + 2).
Restrictions: y ≠ -3, y ≠ 1, y ≠ -2
y  3y  1y  2 
5y(y + 2) – 2(y + 3) = 3y(y – 1)
2

3y
y  2y  3
y  3y  1y  2
Clear the fractions.
2
5y + 10y – 2y – 6 = 3y – 3y
Distribute – watch the minus sign!
2
2y + 11y – 6 = 0
Because the equation is quadratic, set it
equal to 0.
(2y – 1)(y + 6) = 0
Factor
2y – 1 = 0
y=
1
2
Example 4: Solve
y+6=0
Zero Factor Theorem
y = -6
Both are solutions since neither number
is a restriction.
1
2
3


.
a  1 a2  1 a  1
1
2
3


a  1 a  1a  1
a 1
Factor each denominator to find the LCD.
LCD = (a – 1)(a + 1)
Restrictions: a ≠ 1, a ≠ -1
1
2
3
a  1a  1 
a  1a  1 
a  1a  1
a 1
a  1a  1
a1
1(a + 1) – 2 = 3(a – 1)
Reduce to clear fractions.
a + 1 – 2 = 3a – 3
Distribute to remove parentheses
a – 1 = 3a – 3
This is a linear equation so the variable terms
move to one side and constants to the other.
-2a = -2
a=1
V. Zabrocki 2011
1 is an extraneous solution because it makes the
fraction undefined. There is no solution to this
equation. 
page 2
M098
Carson Elementary and Intermediate Algebra 3e
3
Example 5: Solve
2
y 4

2
5y  2
3

3
5y  2y  2 
y  2y  2
y  2y  2
Section 7.6
2
.
5y  10
Factor each denominator to find the LCD.
LCD = 5(y – 2)(y + 2)
Restrictions: y ≠ 2, y ≠ -2
2
5y  2y  2
5y  2
3(5) = -2(y – 2)
Reduce to clear fractions.
15 = -2y + 4
Distribute to remove parentheses
11 = -2y
This is a linear equation so the variable terms move to
one side and constants to the other.
y  
11
2
This is not a restriction.
Example 6: Solve
1
1
6
.
 
2
x3 3
x 9
1
1
6
 
x3 3
x  3x  3
Factor each denominator to find the LCD.
LCD = 3(x – 3)(x + 3)
Restrictions: x ≠ 3, x ≠ -3
1
1
6
3x  3x  3  3x  3x  3 
3x  3x  3
x3
3
x  3x  3
3(x + 3) + (x – 3)(x + 3) = 6(3)
2
3x + 9 + x – 9 = 18
2
x + 3x – 18 = 0
Reduce to clear fractions.
Distribute and foil to remove parentheses
This is a quadratic equation so set it equal to 0.
(x + 6)(x – 3) = 0
Factor and solve.
V. Zabrocki 2011
x+6=0
x–3=0
x = -6
x=3
3 is extraneous so the only solution is -6.
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