Solve for \( y \). \[ -\frac{7}{y+1}=-6-\frac{2}{y-2} \] If there is more than one solution, separate them with commas. If there is no solution, dick on "No solution". \( y=\square \)

To solve for \( y \) in the equation \[ -\frac{7}{y+1}=-6-\frac{2}{y-2}, \] first, eliminate the fractions by multiplying both sides by \((y + 1)(y - 2)\): \[ -7(y - 2) = -6(y + 1)(y - 2) - 2(y + 1). \] Expanding both sides gives: \[ -7y + 14 = -6(y^2 - y - 2) - 2(y + 1). \] Continuing the expansion on the right: \[ -7y + 14 = -6y^2 + 6y + 12 - 2y - 2. \] This simplifies to: \[ -7y + 14 = -6y^2 + 4y + 10. \] Now, rearranging gives: \[ 6y^2 - 7y - 4y + 14 - 10 = 0, \] which simplifies to: \[ 6y^2 - 11y + 4 = 0. \] Next, we can use the quadratic formula where \( a = 6 \), \( b = -11 \), and \( c = 4 \): \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{11 \pm \sqrt{(-11)^2 - 4 \cdot 6 \cdot 4}}{2 \cdot 6}. \] Calculating the discriminant: \[ 121 - 96 = 25. \] Now substituting back into the formula: \[ y = \frac{11 \pm 5}{12}. \] This results in two possible values for \( y \): 1. \( y = \frac{16}{12} = \frac{4}{3} \) 2. \( y = \frac{6}{12} = \frac{1}{2} \) Thus, the solutions for \( y \) are: \[ y = \frac{4}{3}, \frac{1}{2}. \]