1Cademy - Solving $$\frac{x+13}{x^2-7x+10} = \frac{6}{x-5}

Learn Before

Extraneous Solution to a Rational Equation

Example

Solving $\frac{x+13}{x^2-7x+10} = \frac{6}{x-5} - \frac{4}{x-2}$

Solve the rational equation $\frac{x+13}{x^2-7x+10} = \frac{6}{x-5} - \frac{4}{x-2}$ by applying the five-step strategy for equations with rational expressions.

Step 1 — Identify restricted values. Factor the quadratic denominator: $x^2 - 7x + 10 = (x-5)(x-2)$ . Setting each factor equal to zero gives $x = 5$ and $x = 2$ . Record $x eq 5$ and $x eq 2$ .

Step 2 — Find the LCD. The factored denominators are $(x-5)(x-2)$ , $(x-5)$ , and $(x-2)$ . The LCD is $(x-5)(x-2)$ .

Step 3 — Clear the fractions. Multiply every term on both sides by the LCD $(x-5)(x-2)$ and cancel matching denominator factors:

$(x-5)(x-2) \cdot \frac{x+13}{(x-5)(x-2)} = (x-5)(x-2) \cdot \frac{6}{x-5} - (x-5)(x-2) \cdot \frac{4}{x-2}$

Simplify each term to obtain a fraction-free equation:

$x+13 = 6(x-2) - 4(x-5)$

Step 4 — Solve the resulting equation. Distribute on the right: $x+13 = 6x - 12 - 4x + 20$ . Combine like terms: $x+13 = 2x + 8$ . Subtract $x$ from both sides: $13 = x + 8$ . Subtract $8$ from both sides: $x = 5$ .

Step 5 — Check. The algebraic solution $x = 5$ matches one of the restricted values recorded in Step 1, which would make the denominators $(x-5)$ and $(x^2-7x+10)$ equal to zero. Therefore, $x = 5$ is an extraneous solution and must be discarded.

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

Updated 2026-04-30

Contributors are:

Who are from:

References

OpenStax Intermediate Algebra

Learn Before

Related

Learn After