Solve for \( y \). \[ \frac{8}{y^{2}-6 y+8}=\frac{1}{y-4}+\frac{5}{y-2} \] If there is more than one solution, separate them with commas. If there is no solution, click on "No solution". \( y=\square \)

To solve the equation \[ \frac{8}{y^{2}-6 y+8}=\frac{1}{y-4}+\frac{5}{y-2} \] we first simplify the equation by factoring the quadratic in the denominator on the left. The quadratic \( y^{2}-6y+8 \) can be factored as \[ (y-2)(y-4). \] Thus, the equation becomes \[ \frac{8}{(y-2)(y-4)}=\frac{1}{y-4}+\frac{5}{y-2}. \] Next, we find a common denominator for the right side, which is \((y-2)(y-4)\): \[ \frac{1 \cdot (y-2) + 5 \cdot (y-4)}{(y-2)(y-4)}. \] Expanding the numerator, we have \[ (y - 2) + (5y - 20) = 6y - 22. \] Thus, the right side becomes \[ \frac{6y - 22}{(y-2)(y-4)}. \] Now the equation can be rewritten as \[ \frac{8}{(y-2)(y-4)}=\frac{6y - 22}{(y-2)(y-4)}. \] Since the denominators are equal and assuming \(y \neq 2\) and \(y \neq 4\), we can equate the numerators: \[ 8 = 6y - 22. \] Rearranging this gives \[ 6y = 30 \implies y = 5. \] Finally, we check the restrictions. Since \(y\) must not equal 2 or 4 to avoid division by zero, and 5 is not in that set, we conclude that the only solution is \[ y=5. \] Thus, the answer is \( y=5 \).