1Cademy - Solving $$\frac{4}{q-4}

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Strategy to Solve Equations with Rational Expressions

Example

Solving $\frac{4}{q-4} - \frac{3}{q-3} = 1$

Solve the rational equation $\frac{4}{q-4} - \frac{3}{q-3} = 1$ by applying the five-step strategy for equations with rational expressions. This example demonstrates a case where clearing fractions produces a quadratic equation that requires factoring and the Zero Product Property.

Step 1 — Identify restricted values. Setting each denominator equal to zero: $q - 4 = 0$ gives $q = 4$ , and $q - 3 = 0$ gives $q = 3$ . Record $q \neq 4$ and $q \neq 3$ .

Step 2 — Find the LCD. The denominators $(q-4)$ and $(q-3)$ share no common factors, and the right side has an implicit denominator of $1$ . The LCD is $(q-4)(q-3)$ .

Step 3 — Clear the fractions. Multiply both sides by the LCD $(q-4)(q-3)$ and distribute. Cancel matching denominator factors: the first term becomes $4(q-3)$ , the second becomes $3(q-4)$ , and the right side becomes $(q-4)(q-3)$ :

$4(q-3) - 3(q-4) = (q-4)(q-3)$

Step 4 — Solve the resulting equation. Distribute on the left: $4q - 12 - 3q + 12 = q^2 - 7q + 12$ . Combine like terms on the left: $q = q^2 - 7q + 12$ . Write in standard form by subtracting $q$ from both sides: $0 = q^2 - 8q + 12$ . Factor the trinomial: $0 = (q - 2)(q - 6)$ . Apply the Zero Product Property:

$q - 2 = 0 \quad \text{or} \quad q - 6 = 0$

$q = 2 \quad \text{or} \quad q = 6$

Step 5 — Check. Neither $2$ nor $6$ equals a restricted value ( $4$ or $3$ ), so neither is extraneous.

For $q = 2$ : $\frac{4}{2-4} - \frac{3}{2-3} = \frac{4}{-2} - \frac{3}{-1} = -2 - (-3) = -2 + 3 = 1$ ✓

For $q = 6$ : $\frac{4}{6-4} - \frac{3}{6-3} = \frac{4}{2} - \frac{3}{3} = 2 - 1 = 1$ ✓

The solutions are $q = 2$ and $q = 6$ . This equation has the integer $1$ on the right side and two distinct linear binomial denominators on the left. Multiplying both sides by the LCD $(q-4)(q-3)$ produces a quadratic on the right side — because the LCD is itself a product of two binomials — while the left side simplifies to a single linear term. The resulting quadratic $q^2 - 8q + 12 = 0$ must be solved by factoring and the Zero Product Property, yielding two valid solutions.

Updated 2026-04-21

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OpenStax Elementary Algebra

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