Solve for \( y \). \[ \frac{1}{y-4}+\frac{2}{y+3}=\frac{10}{y^{2}-y-12} \] If there is more than one solution, separate them with commas. If there is no solution, dick on "No solution". \( y=\square \)

To solve the equation: \[ \frac{1}{y-4} + \frac{2}{y+3} = \frac{10}{y^{2} - y - 12} \] **Step 1:** Factor the denominator on the right side. \[ y^{2} - y - 12 = (y - 4)(y + 3) \] **Step 2:** Rewrite the equation with the factored denominator. \[ \frac{1}{y-4} + \frac{2}{y+3} = \frac{10}{(y-4)(y+3)} \] **Step 3:** Multiply both sides by \((y - 4)(y + 3)\) to eliminate the denominators. \[ (y + 3) + 2(y - 4) = 10 \] **Step 4:** Distribute and combine like terms. \[ y + 3 + 2y - 8 = 10 \] \[ 3y - 5 = 10 \] **Step 5:** Solve for \( y \). \[ 3y = 15 \] \[ y = 5 \] **Step 6:** Check for any restrictions. The original denominators \( y - 4 \) and \( y + 3 \) cannot be zero. \[ y \neq 4 \quad \text{and} \quad y \neq -3 \] Since \( y = 5 \) does not violate these restrictions, it is the valid solution. \[ y = 5 \]