Question

Please help I’m struggling on this question.

Answer 1

Answer: $\frac{2(y^2-4y+40)}{(y-8)(y+4)}$

Step-by-step explanation:

To solve this problem, we want to make sure the denominators are the same on both fractions. Once they are equal, we can dd them together.

$\frac{y+4}{y-8} +\frac{y-8}{y+4}$                      [multiply first fraction by y+4 and second by y-8]

$\frac{(y+4)(y+4)}{(y-8)(y+4)} +\frac{(y-8)(y-8)}{(y-8)(y+4)}$   [distribute by FOIL]

$\frac{y^2+8y+16}{(y-8)(y+4)} +\frac{y^2-16y+64}{(y-8)(y+4)}$   [add numerator]

$\frac{y^2+8y+16+y^2-16y+64}{(y-8)(y+4)}$         [combine like terms]

$\frac{2y^2-8y+80}{(y-8)(y+4)}$                       [factor out 2 from numerator]

$\frac{2(y^2-4y+40)}{(y-8)(y+4)}$                  

Now we know that $\frac{2(y^2-4y+40)}{(y-8)(y+4)}$ is the factored form after adding.