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We always start by noting the values that would cause any denominators to be zero.

Solve: 1 5 y = 6 y 2 .

Solution

.
Note any value of the variable that would make any denominator zero. .
Find the least common denominator of all denominators in the equation. The LCD is y 2 .
Clear the fractions by multiplying both sides of the equation by the LCD. .
Distribute. .
Multiply. .
Solve the resulting equation. First write the quadratic equation in standard form. .
Factor. .
Use the Zero Product Property. .
Solve. .
Check.
We did not get 0 as an algebraic solution.
.

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Solve: 1 2 a = 15 a 2 .

5 , −3

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Solve: 1 4 b = 12 b 2 .

6 , −2

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Solve: 5 3 u 2 = 3 2 u .

Solution

.
Note any value of the variable that would make any denominator zero. .
Find the least common denominator of all denominators in the equation. The LCD is 2 u ( 3 u 2 ) .
Clear the fractions by multiplying both sides of the equation by the LCD. .
Remove common factors. .
Simplify. .
Multiply. .
Solve the resulting equation. .
We did not get 0 or 2 3 as algebraic solutions.
.

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Solve: 1 x 1 = 2 3 x .

−2

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Solve: 3 5 n + 1 = 2 3 n .

−2

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When one of the denominators is a quadratic, remember to factor it first to find the LCD.

Solve: 2 p + 2 + 4 p 2 = p 1 p 2 4 .

Solution

.
Note any value of the variable that would make any denominator zero. .
Find the least common denominator of all denominators in the equation. The LCD is ( p + 2 ) ( p 2 ) .
Clear the fractions by multiplying both sides of the equation by the LCD. .
Distribute. .
Remove common factors. .
Simplify. .
Distribute. .
Solve. .
.
.
We did not get 2 or 2 as algebraic solutions.
.

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Solve: 2 x + 1 + 1 x 1 = 1 x 2 1 .

2 3

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Solve: 5 y + 3 + 2 y 3 = 5 y 2 9 .

2

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Solve: 4 q 4 3 q 3 = 1 .

Solution

.
Note any value of the variable that would make any denominator zero. .
Find the least common denominator of all denominators in the equation. The LCD is ( q 4 ) ( q 3 ) .
Clear the fractions by multiplying both sides of the equation by the LCD. .
Distribute. .
Remove common factors. .
Simplify. .
Simplify. .
Combine like terms. .
Solve. First write in standard form. .
Factor. .
Use the Zero Product Property. .
We did not get 4 or 3 as algebraic solutions.
.

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Solve: 2 x + 5 1 x 1 = 1 .

−1 , −2

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Solve: 3 x + 8 2 x 2 = 1 .

−2 , −3

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Solve: m + 11 m 2 5 m + 4 = 5 m 4 3 m 1 .

Solution

.
Factor all the denominators, so we can note any value of the variable the would make any denominator zero. .
Find the least common denominator of all denominators in the equation. The LCD is ( m 4 ) ( m 1 ) .
Clear the fractions. .
Distribute. .
Remove common factors. .
Simplify. .
Solve the resulting equation. .
.
Check. The only algebraic solution was 4, but we said that 4 would make a denominator equal to zero. The algebraic solution is an extraneous solution. There is no solution to this equation.

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Solve: x + 13 x 2 7 x + 10 = 6 x 5 4 x 2 .

no solution

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Solve: y 14 y 2 + 3 y 4 = 2 y + 4 + 7 y 1 .

no solution

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The equation we solved in [link] had only one algebraic solution, but it was an extraneous solution. That left us with no solution to the equation. Some equations have no solution.

Solve: n 12 + n + 3 3 n = 1 n .

Solution

.
Note any value of the variable that would make any denominator zero. .
Find the least common denominator of all denominators in the equation. The LCD is 12 n .
Clear the fractions by multiplying both sides of the equation by the LCD. .
Distribute. .
Remove common factors. .
Simplify. .
Solve the resulting equation. .
.
.
.
Check.
n = 0 is an extraneous solution.
.

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Practice Key Terms 2

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Source:  OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2
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