Solve the rational equation: $ \frac{x+1}{x-2} - \frac{3}{x-4} = \frac{x^2 - 9x + 5}{(x+1)(x-4)} $ a. \(...

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12. Solve the rational equation: $ \frac{x+1}{x-2} - \frac{3}{x-4} = \frac{x^2 - 9x + 5}{(x+1)(x-4)} $ a. $ x = 4 $ b. $ x = \frac{1}{2} $ c. No solution d. $ x = \frac{1}{2}, 4 $

SnapAndSolve AI · Accepted Answer

To solve the rational equation $ \frac{x+1}{x-2} - \frac{3}{x-4} = \frac{x^2 - 9x + 5}{(x+1)(x-4)} $, we first find a common denominator, which is $(x-2)(x-4)(x+1)$. Multiply each term by the common denominator to eliminate the fractions: $(x+1)(x-4)(x+1) - 3(x-2)(x+1) = (x^2 - 9x + 5)(x-2)$ Expand and simplify each side: $(x^2 - 3x - 4)(x+1) - 3(x^2 - 2x) = x^3 - 11x^2 + 23x - 10$ $x^3 - 3x^2 - 4x + x^2 - 3x - 4 - 3x^2 + 6x = x^3 - 11x^2 + 23x - 10$ Simplify further: $x^3 - 5x^2 - x - 4 = x^3 - 11x^2 + 23x - 10$ Cancel $x^3$ from both sides: $-5x^2 - x - 4 = -11x^2 + 23x - 10$ Rearrange terms: $6x^2 - 24x + 6 = 0$ Divide the entire equation by 6: $x^2 - 4x + 1 = 0$ Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{4 \pm \sqrt{16 - 4}}{2}$ $x = \frac{4 \pm \sqrt{12}}{2}$ $x = \frac{4 \pm 2\sqrt{3}}{2}$ $x = 2 \pm \sqrt{3}$ Check for extraneous solutions by substituting back into the original equation. Neither $x = 2 + \sqrt{3}$ nor $x = 2 - \sqrt{3}$ are valid due to restrictions from the original denominators. Thus, the correct answer is c. No solution.

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Solve the rational equation: \( \frac{x+1}{x-2} - \frac{3}{x-4} = \frac{x^2 - 9x + 5}{(x+1)(x-4)} \) a. \(...

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