Solve for \( y \). \( \frac{6}{y-5}+\frac{y}{y-3}=\frac{24}{y^{2}-8 y+15} \) If there is more than one solution, separate them with commas. If there is no solution, click on "No solution".

To solve the equation \( \frac{6}{y-5}+\frac{y}{y-3}=\frac{24}{y^{2}-8y+15} \), we first notice that the denominator on the right side can be factored. The quadratic \( y^{2}-8y+15 \) factors to \( (y-5)(y-3) \). This allows us to rewrite the equation as follows: \[ \frac{6}{y-5}+\frac{y}{y-3}=\frac{24}{(y-5)(y-3)}. \] Next, we can get a common denominator for the left side of the equation, which is \( (y-5)(y-3) \): \[ \frac{6(y-3)}{(y-5)(y-3)} + \frac{y(y-5)}{(y-5)(y-3)} = \frac{24}{(y-5)(y-3)}. \] Combining the fractions on the left gives us: \[ \frac{6(y-3) + y(y-5)}{(y-5)(y-3)} = \frac{24}{(y-5)(y-3)}. \] Now, cross-multiplying leads to: \[ 6(y-3) + y(y-5) = 24. \] Expanding both sides: \[ 6y - 18 + y^2 - 5y = 24. \] Combining like terms results in: \[ y^2 + y - 18 = 24. \] Moving all terms to one side gives: \[ y^2 + y - 42 = 0. \] Next, we factor this quadratic: \[ (y - 6)(y + 7) = 0. \] Thus, the solutions for \( y \) are: \[ y - 6 = 0 \quad \Rightarrow \quad y = 6, \] \[ y + 7 = 0 \quad \Rightarrow \quad y = -7. \] Now, let's check if the potential solutions \( y=6 \) and \( y=-7 \) don't make any denominators zero. For \( y-5 \) and \( y-3 \): - For \( y=6 \): \( 6-5=1 \neq 0\) and \( 6-3=3 \neq 0\). - For \( y=-7 \): \( -7-5=-12 \neq 0\) and \( -7-3=-10 \neq 0\). Both values are valid solutions since they don't cause any denominator to be zero. Thus, the final solutions are: \[ \boxed{6, -7}. \]