1Cademy - Solving $$\frac{x}{2x-2} - \frac{2}{3x+3} = \frac{5x^2-2x+9}{12x^2-12}$$

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Strategy to Solve Equations with Rational Expressions

Example

Solving $\frac{x}{2x-2} - \frac{2}{3x+3} = \frac{5x^2-2x+9}{12x^2-12}$

Solve the rational equation $\frac{x}{2x-2} - \frac{2}{3x+3} = \frac{5x^2-2x+9}{12x^2-12}$ by applying the five-step strategy for equations with rational expressions. This example demonstrates a case where every algebraic solution turns out to be extraneous, so the equation has no solution.

Step 1 — Factor all denominators and identify restricted values. Factor each denominator to make the LCD and restricted values easier to find:

$2x - 2 = 2(x-1)$
$3x + 3 = 3(x+1)$
$12x^2 - 12 = 12(x^2 - 1) = 12(x-1)(x+1)$

Setting each variable factor equal to zero: $x - 1 = 0$ gives $x = 1$ , and $x + 1 = 0$ gives $x = -1$ . Record $x \neq 1$ and $x \neq -1$ .

Step 2 — Find the LCD. The factored denominators are $2(x-1)$ , $3(x+1)$ , and $12(x-1)(x+1)$ . The LCD must include the highest power of every factor: $12(x-1)(x+1)$ .

Step 3 — Clear the fractions. Multiply every term by the LCD $12(x-1)(x+1)$ and cancel matching factors:

$12(x-1)(x+1) \cdot \frac{x}{2(x-1)} - 12(x-1)(x+1) \cdot \frac{2}{3(x+1)} = 12(x-1)(x+1) \cdot \frac{5x^2-2x+9}{12(x-1)(x+1)}$

Simplify: the first term becomes $6(x+1) \cdot x = 6x(x+1)$ , the second becomes $4(x-1) \cdot 2 = 8(x-1)$ , and the right side becomes $5x^2 - 2x + 9$ :

$6x(x+1) - 8(x-1) = 5x^2 - 2x + 9$

Step 4 — Solve the resulting equation. Distribute: $6x^2 + 6x - 8x + 8 = 5x^2 - 2x + 9$ . Combine like terms on the left: $6x^2 - 2x + 8 = 5x^2 - 2x + 9$ . Subtract $5x^2 - 2x + 9$ from both sides: $x^2 - 1 = 0$ . Factor as a difference of squares: $(x-1)(x+1) = 0$ . Apply the Zero Product Property:

$x - 1 = 0 \quad \text{or} \quad x + 1 = 0$

$x = 1 \quad \text{or} \quad x = -1$

Step 5 — Check. Both $x = 1$ and $x = -1$ are restricted values identified in Step 1. Substituting either value into the original equation would produce a zero denominator, making the equation undefined. Therefore, both are extraneous solutions.

The equation has no solution.

This example is the most extreme case of extraneous solutions: the algebraic process produces candidates, but every single one matches a restricted value. When all solutions of a rational equation are extraneous, the original equation has no valid solution. This outcome is possible whenever the resulting polynomial equation after clearing fractions happens to have roots that coincide exactly with the values excluded in Step 1.

Updated 2026-04-21

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OpenStax Elementary Algebra

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