To solve the equation [tex]\(\frac{-8}{2 y-8}=\frac{5}{y+4}-\frac{7 y+8}{y^2-16}\)[/tex], let's go through it step-by-step.
First, simplify and rewrite the terms wherever possible:
1. Recall that:
[tex]\[
y^2 - 16 = (y - 4)(y + 4)
\][/tex]
2. Rewrite the original equation:
[tex]\[
\frac{-8}{2 y-8}=\frac{5}{y+4}-\frac{7 y+8}{(y-4)(y+4)}
\][/tex]
3. Simplify the first fraction on the left-hand side:
[tex]\[
2y - 8 = 2(y - 4) \implies \frac{-8}{2(y-4)} = \frac{-4}{y-4}
\][/tex]
So, the equation becomes:
[tex]\[
\frac{-4}{y-4} = \frac{5}{y+4} - \frac{7y+8}{(y-4)(y+4)}
\][/tex]
4. To have a common denominator, combine the right-hand side terms under a single denominator:
[tex]\[
\frac{5}{y+4} - \frac{7y + 8}{(y-4)(y+4)}
\][/tex]
5. Express each fraction with the common denominator [tex]\((y-4)(y+4)\)[/tex]:
[tex]\[
\frac{5(y-4)}{(y-4)(y+4)} - \frac{7y+8}{(y-4)(y+4)}
\][/tex]
6. Combine the fractions:
[tex]\[
\frac{5(y-4) - (7y+8)}{(y-4)(y+4)}
\][/tex]
7. Simplify the numerator:
[tex]\[
5(y-4) = 5y - 20
\][/tex]
[tex]\[
5(y-4) - (7y+8) = 5y - 20 - 7y - 8 = -2y - 28
\][/tex]
8. This gives:
[tex]\[
\frac{-2y - 28}{(y-4)(y+4)}
\][/tex]
So, the equation now looks like:
[tex]\[
\frac{-4}{y-4} = \frac{-2y - 28}{(y-4)(y+4)}
\][/tex]
9. Cross-multiply to eliminate the fractions:
[tex]\[
-4(y+4) = (-2y-28)
\][/tex]
10. Simplify:
[tex]\[
-4y - 16 = -2y - 28
\][/tex]
11. Bring like terms together:
[tex]\[
-4y + 2y = -28 + 16
\][/tex]
[tex]\[
-2y = -12
\][/tex]
12. Solve for [tex]\( y \)[/tex]:
[tex]\[
y = 6
\][/tex]
Therefore, the solution to the equation is:
[tex]\[
y = 6
\][/tex]
