Example

Solving 43x2−10x+3+33x2+2x−1=2x2−2x−3\frac{4}{3x^2-10x+3} + \frac{3}{3x^2+2x-1} = \frac{2}{x^2-2x-3}

Solve the rational equation 43x2−10x+3+33x2+2x−1=2x2−2x−3\frac{4}{3x^2-10x+3} + \frac{3}{3x^2+2x-1} = \frac{2}{x^2-2x-3} by applying the five-step strategy for equations with rational expressions.

Step 1 — Identify restricted values. Factor all three denominators: 3x2−10x+3=(3x−1)(x−3)3x^2-10x+3 = (3x-1)(x-3) 3x2+2x−1=(3x−1)(x+1)3x^2+2x-1 = (3x-1)(x+1) x2−2x−3=(x−3)(x+1)x^2-2x-3 = (x-3)(x+1) Setting each unique factor to zero gives the restricted values: x≠−1x \neq -1, x≠13x \neq \frac{1}{3}, and x≠3x \neq 3.

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

Step 3 — Clear the fractions. Multiply every term on both sides by the LCD (3x−1)(x+1)(x−3)(3x-1)(x+1)(x-3) and cancel matching denominator factors: (3x−1)(x+1)(x−3)⋅4(3x−1)(x−3)+(3x−1)(x+1)(x−3)⋅3(3x−1)(x+1)=(3x−1)(x+1)(x−3)⋅2(x−3)(x+1)(3x-1)(x+1)(x-3) \cdot \frac{4}{(3x-1)(x-3)} + (3x-1)(x+1)(x-3) \cdot \frac{3}{(3x-1)(x+1)} = (3x-1)(x+1)(x-3) \cdot \frac{2}{(x-3)(x+1)} Simplify each term to obtain a fraction-free equation: 4(x+1)+3(x−3)=2(3x−1)4(x+1) + 3(x-3) = 2(3x-1)

Step 4 — Solve the resulting equation. Distribute: 4x+4+3x−9=6x−24x + 4 + 3x - 9 = 6x - 2. Combine like terms: 7x−5=6x−27x - 5 = 6x - 2. Subtract 6x6x from both sides: x−5=−2x - 5 = -2. Add 55 to both sides: x=3x = 3.

Step 5 — Check. The algebraic solution x=3x = 3 exactly matches one of the restricted values identified in Step 1. Substituting x=3x = 3 into the original equation would make two of the denominators equal to zero, which is mathematically undefined. Therefore, x=3x = 3 is an extraneous solution and must be discarded.

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

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Updated 2026-06-03

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