Example

Solving m+11m25m+4=5m43m1\frac{m+11}{m^2-5m+4} = \frac{5}{m-4} - \frac{3}{m-1}

Solve the rational equation m+11m25m+4=5m43m1\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: m25m+4=(m4)(m1)m^2 - 5m + 4 = (m-4)(m-1). Setting each factor equal to zero gives m=4m = 4 and m=1m = 1. Record meq4m eq 4 and meq1m eq 1.

Step 2 — Find the LCD. The factored denominators are (m4)(m1)(m-4)(m-1), (m4)(m-4), and (m1)(m-1). The LCD is (m4)(m1)(m-4)(m-1).

Step 3 — Clear the fractions. Multiply every term on both sides by the LCD (m4)(m1)(m-4)(m-1) and cancel matching denominator factors:

(m4)(m1)m+11(m4)(m1)=(m4)(m1)5m4(m4)(m1)3m1(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+11m+11. On the right, the first term becomes 5(m1)5(m-1), and the second term becomes 3(m4)3(m-4):

m+11=5(m1)3(m4)m+11 = 5(m-1) - 3(m-4)

Step 4 — Solve the resulting equation. Distribute on the right: m+11=5m53m+12m+11 = 5m - 5 - 3m + 12. Combine like terms: m+11=2m+7m+11 = 2m + 7. Subtract mm from both sides: 11=m+711 = m + 7. Subtract 77 from both sides: 4=m4 = m, so m=4m = 4.

Step 5 — Check. The algebraic solution m=4m = 4 matches one of the restricted values recorded in Step 1. Substituting m=4m = 4 would make the denominators (m4)(m-4) and (m25m+4)(m^2-5m+4) equal zero, which is undefined. Therefore, m=4m = 4 is an extraneous solution and must be discarded.

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

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Updated 2026-05-25

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