1Cademy - Solving $$\frac{5}{x^2+2x-3} - \frac{3}{x^2+x-2} = \frac{1}{x^2+5x+6}$$

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

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Solving $\frac{5}{x^2+2x-3} - \frac{3}{x^2+x-2} = \frac{1}{x^2+5x+6}$

Solve the rational equation $\frac{5}{x^2+2x-3} - \frac{3}{x^2+x-2} = \frac{1}{x^2+5x+6}$ .

Step 1 — Identify restricted values. Factor all three denominators: $x^2+2x-3 = (x+3)(x-1)$ $x^2+x-2 = (x+2)(x-1)$ $x^2+5x+6 = (x+2)(x+3)$ Setting each unique factor to zero gives the restricted values: $x \neq -3$ , $x \neq 1$ , and $x \neq -2$ .

Step 2 — Find the LCD. The factored denominators share the factors $(x+3)$ , $(x-1)$ , and $(x+2)$ . The LCD is $(x+3)(x-1)(x+2)$ .

Step 3 — Clear the fractions. Multiply every term by the LCD $(x+3)(x-1)(x+2)$ and cancel matching denominator factors: $(x+3)(x-1)(x+2) \cdot \frac{5}{(x+3)(x-1)} - (x+3)(x-1)(x+2) \cdot \frac{3}{(x+2)(x-1)} = (x+3)(x-1)(x+2) \cdot \frac{1}{(x+2)(x+3)}$ Simplify to obtain the fraction-free equation: $5(x+2) - 3(x+3) = 1(x-1)$

Step 4 — Solve the resulting equation. Distribute: $5x + 10 - 3x - 9 = x - 1$ . Combine like terms: $2x + 1 = x - 1$ . Subtract $x$ from both sides: $x + 1 = -1$ . Subtract $1$ from both sides: $x = -2$ .

Step 5 — Check. The algebraic solution $x = -2$ matches one of the restricted values. Because substituting $x = -2$ would make two denominators equal to zero, it is an extraneous solution.

Since the algebraic process yielded only an extraneous solution, there is no solution to this equation.

Updated 2026-05-25

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OpenStax Intermediate Algebra

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