Carson Elementary And Intermediate Algebra 3e Page 2
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1
2
3M098
Carson Elementary and Intermediate Algebra 3e
Section 7.6
3
x
2
Example 2: What are the restrictions for
7
x
?
5
3
There are no restrictions because there are no variables in the denominator. The
fractions will never be undefined.
5
y
2
3
y
Example 3: Solve
.
2
2
2
y
2
y
3
y
y
2
y
5
y
6
5
y
2
3
y
Factor each denominator to find the LCD.
LCD = (y + 3)(y – 1)(y + 2).
y
3
y
1
y
2
y
1
y
2
y
3
Restrictions: y ≠ -3, y ≠ 1, y ≠ -2
5
y
2
3
y
y
3
y
1
y
2
y
3
y
1
y
2
y
3
y
1
y
2
y
3
y
1
y
2
y
1
y
2
y
3
5y(y + 2) – 2(y + 3) = 3y(y – 1)
Clear the fractions.
2
2
+ 10y – 2y – 6 = 3y
– 3y
Distribute – watch the minus sign!
5y
2
+ 11y – 6 = 0
2y
Because the equation is quadratic, set it
equal to 0.
(2y – 1)(y + 6) = 0
Factor
2y – 1 = 0
y + 6 = 0
Zero Factor Theorem
1
Both are solutions since neither number
y =
y = -6
is a restriction.
2
1
2
3
Example 4: Solve
.
2
a
1
a
1
a
1
1
2
3
Factor each denominator to find the LCD.
LCD = (a – 1)(a + 1)
a
1
a
1
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
1 is an extraneous solution because it makes the
fraction undefined. There is no solution to this
equation.
V. Zabrocki 2011
page 2
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