Question

going to ask this again, please I need help, this is my last one for tonight.

4x-8/y+2 divided by y-2/y^2-4

Answer 1

I assume this is the problem:
[(4x-8)/(y+2)]/[(y-2)/(y^2-4)]

[(4x-8)/(y+2)][(y^2-4)/(y-2)]

[(4x-8)/(y+2)][(y-2)(y+2)/(y-2)]

The answer is: 4x-8 = 4(x-2)

Answer 2

\[\left(\begin{matrix}4x-8 \\ y+2\end{matrix}\right)\div \left(\begin{matrix}y-2 \\ y ^{2}-4\end{matrix}\right)\]

Answer 3

O.K. The answer I gave is for the problem you just stated.

Answer 4

ok thank you :)

Answer 5

You're welcome