4) \( \frac{4 x-8}{x^{2}-x-2}=3 \)

To solve the equation \( \frac{4 x-8}{x^{2}-x-2}=3 \), start by eliminating the fraction. Multiply both sides by \( x^{2} - x - 2 \) (assuming \( x^{2} - x - 2 \neq 0 \)): \[ 4x - 8 = 3(x^{2} - x - 2) \] This expands to: \[ 4x - 8 = 3x^{2} - 3x - 6 \] Rearranging gives: \[ 3x^{2} - 7x + 2 = 0 \] Now, use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 3, b = -7, c = 2 \). Calculating the discriminant: \[ b^2 - 4ac = (-7)^2 - 4(3)(2) = 49 - 24 = 25 \] Now substitute into the quadratic formula: \[ x = \frac{7 \pm \sqrt{25}}{2(3)} = \frac{7 \pm 5}{6} \] This gives: \[ x = \frac{12}{6} = 2 \quad \text{and} \quad x = \frac{2}{6} = \frac{1}{3} \] The solutions are \( x = 2 \) and \( x = \frac{1}{3} \). Remember to check for extraneous solutions by substituting back into the original equation. Both values hold true, so your final answers are: **\( x = 2 \) and \( x = \frac{1}{3} \)**.