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

To solve the equation \[ -\frac{1}{w+2}=-4+\frac{3}{w+3} \] First, we can isolate the fractions on the right by writing it as: \[ -\frac{1}{w+2} + 4 = \frac{3}{w+3} \] Next, multiply through by the common denominator \( (w+2)(w+3) \) to eliminate the fractions: \[ -1(w+3) + 4(w+2)(w+3) = 3(w+2) \] Expanding all terms will give: \[ -w - 3 + 4(w^2 + 5w + 6) = 3w + 6 \] which simplifies to: \[ -w - 3 + 4w^2 + 20w + 24 = 3w + 6 \] Now, combining like terms: \[ 4w^2 + 19w + 21 = 3w + 6 \] This leads us to: \[ 4w^2 + 16w + 15 = 0 \] Next, we factor this quadratic equation: \[ (4w + 3)(w + 5) = 0 \] Setting each factor to zero gives: \[ 4w + 3 = 0 \quad \text{or} \quad w + 5 = 0 \] From \( 4w + 3 = 0 \): \[ w = -\frac{3}{4} \] From \( w + 5 = 0 \): \[ w = -5 \] Thus, the solutions are: \( w = -\frac{3}{4}, -5 \)