1Cademy - Solving $$\frac{m+11}{m^2-5m+4} = \frac{5}{m-4}

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Extraneous Solution to a Rational Equation

Example

Solving $\frac{m+11}{m^2-5m+4} = \frac{5}{m-4} - \frac{3}{m-1}$

Solve the rational equation $\frac{m+11}{m^2-5m+4} = \frac{5}{m-4} - \frac{3}{m-1}$ by applying the five-step strategy for equations with rational expressions. This example demonstrates a case where the only algebraic solution is extraneous, resulting in no valid solution for the equation.

Step 1 — Identify restricted values. Factor the quadratic denominator: $m^2 - 5m + 4 = (m-4)(m-1)$ . Setting each factor equal to zero gives $m = 4$ and $m = 1$ . Record $m eq 4$ and $m eq 1$ .

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

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

$(m-4)(m-1) \cdot \frac{m+11}{(m-4)(m-1)} = (m-4)(m-1) \cdot \frac{5}{m-4} - (m-4)(m-1) \cdot \frac{3}{m-1}$

Simplify each term: on the left, the entire denominator cancels, leaving $m+11$ . On the right, the first term becomes $5(m-1)$ , and the second term becomes $3(m-4)$ :

$m+11 = 5(m-1) - 3(m-4)$

Step 4 — Solve the resulting equation. Distribute on the right: $m+11 = 5m - 5 - 3m + 12$ . Combine like terms: $m+11 = 2m + 7$ . Subtract $m$ from both sides: $11 = m + 7$ . Subtract $7$ from both sides: $4 = m$ , so $m = 4$ .

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

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

Updated 2026-04-30

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OpenStax Intermediate Algebra

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