Solve the equation. \( \frac{4}{w^{2}+2 w-8}+\frac{3}{w^{2}+w-6}=\frac{4}{w^{2}+7 w+12} \)

To solve the equation: \[ \frac{4}{w^{2}+2 w-8} + \frac{3}{w^{2}+w-6} = \frac{4}{w^{2}+7 w+12} \] **Step 1: Factor the Denominators** First, factor each quadratic expression in the denominators: 1. \( w^{2} + 2w - 8 \): \[ w^{2} + 2w - 8 = (w + 4)(w - 2) \] 2. \( w^{2} + w - 6 \): \[ w^{2} + w - 6 = (w + 3)(w - 2) \] 3. \( w^{2} + 7w + 12 \): \[ w^{2} + 7w + 12 = (w + 3)(w + 4) \] **Step 2: Rewrite the Equation with Factored Denominators** Substituting the factored forms into the original equation: \[ \frac{4}{(w + 4)(w - 2)} + \frac{3}{(w + 3)(w - 2)} = \frac{4}{(w + 3)(w + 4)} \] **Step 3: Find the Least Common Denominator (LCD)** The LCD of the denominators is \((w + 4)(w + 3)(w - 2)\). Multiply both sides of the equation by this LCD to eliminate the denominators: \[ 4(w + 3) + 3(w + 4) = 4(w - 2) \] **Step 4: Simplify and Solve for \( w \)** Expanding and simplifying: \[ 4(w + 3) + 3(w + 4) = 4w - 8 \\ 4w + 12 + 3w + 12 = 4w - 8 \\ 7w + 24 = 4w - 8 \\ 7w - 4w = -8 - 24 \\ 3w = -32 \\ w = -\frac{32}{3} \] **Step 5: Verify the Solution** Ensure that \( w = -\frac{32}{3} \) does not make any denominator zero: - \( w + 4 = -\frac{32}{3} + 4 = -\frac{32}{3} + \frac{12}{3} = -\frac{20}{3} \neq 0 \) - \( w + 3 = -\frac{32}{3} + 3 = -\frac{32}{3} + \frac{9}{3} = -\frac{23}{3} \neq 0 \) - \( w - 2 = -\frac{32}{3} - 2 = -\frac{32}{3} - \frac{6}{3} = -\frac{38}{3} \neq 0 \) Since none of the denominators are zero, the solution is valid. **Final Answer:** \[ w = -\frac{32}{3} \]