1Cademy - Solving $$\frac{y-6}{y^2+3y-4} = \frac{2}{y+4} + \frac{7}{y-1}$$

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Extraneous Solution to a Rational Equation

Example

Solving $\frac{y-6}{y^2+3y-4} = \frac{2}{y+4} + \frac{7}{y-1}$

Solve the rational equation $\frac{y-6}{y^2+3y-4} = \frac{2}{y+4} + \frac{7}{y-1}$ by applying the five-step strategy for equations with rational expressions.

Step 1 — Identify restricted values. Factor the quadratic denominator: $y^2 + 3y - 4 = (y+4)(y-1)$ . Setting each factor equal to zero gives $y = -4$ and $y = 1$ . Record $y eq -4$ and $y eq 1$ .

Step 2 — Find the LCD. The factored denominators are $(y+4)(y-1)$ , $(y+4)$ , and $(y-1)$ . The LCD is $(y+4)(y-1)$ .

Step 3 — Clear the fractions. Multiply every term on both sides by the LCD $(y+4)(y-1)$ and cancel matching denominator factors:

$(y+4)(y-1) \cdot \frac{y-6}{(y+4)(y-1)} = (y+4)(y-1) \cdot \frac{2}{y+4} + (y+4)(y-1) \cdot \frac{7}{y-1}$

Simplify each term to obtain a fraction-free equation:

$y-6 = 2(y-1) + 7(y+4)$

Step 4 — Solve the resulting equation. Distribute on the right: $y-6 = 2y - 2 + 7y + 28$ . Combine like terms: $y-6 = 9y + 26$ . Subtract $y$ from both sides: $-6 = 8y + 26$ . Subtract $26$ from both sides: $-32 = 8y$ . Divide by $8$ : $y = -4$ .

Step 5 — Check. The algebraic solution $y = -4$ matches one of the restricted values recorded in Step 1, which would make the denominators $(y+4)$ and $(y^2+3y-4)$ equal to zero. Therefore, $y = -4$ is an extraneous solution and must be discarded.

Because the only candidate solution is extraneous, the original equation has no solution.

Updated 2026-06-05

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OpenStax Intermediate Algebra

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