PLEASE HELP ME!!! Given the equation below, solve for y.

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PLEASE HELP ME!!! Given the equation below, solve for y.

Mathematics
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z-4=\[z-4\frac{ y ^{2}+8z-48}{ z-4 }\]
@Jhannybean please help me
@kohai please can you help me??

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you get anyqwhere?
@DanJS please help me
is it going to be y=z-4?
you need to isolate y on one side and it will = something in terms of all the other stuff, x
I would multiply both sides of the thing by (z-4) first
y=z-8 is this the answer?
should that be a +8y instead of the +8z you typed?
its a multiple choice question is my answer right?
|dw:1443160932457:dw|
its 8z
ok, just checking, that makes it easier, asking because a y would be a factorable numorator.
multiply both sides by (z-4) (z - 4)*(z-4) = y^2 + 8z - 48
expand z^2 - 8z + 16 = y^2 + 8z - 48
\[y=\sqrt{z ^{2}+6z-40}\]
move the 8z and the -48 to the other side... z^2 - 16z + 64 = y^2
take the root of both sides, and you get two answers for y, the + and the - root of all that on the left
\[y = \pm \sqrt{z^2-16z+64}\]
looking closer, the stuff under the root is a perfect square, factors to (z-8)^2 so you get simplified y = z - 8 or y = -(z-8)
so thats the answer? because i said y=z-8
sorry, back
yeah if that is the only choice in the multiple choice , there is also the negative of that , and there is a restriction on the values of z, but not sur eif you need that

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