Please ckeck?

Question

2y/y+4 -2y/y-4 = y^2+16/y^2-16

2y(y-4) – 2y(y+4) = y^2+16/(y+4)(y-4)

2y^2-8y-2y^2-8y = y^2+16

-16y = y^2+16

0 = y^2+16 + 16

Answer 1

((2y)/(y + 4)) - ((2y)/(y - 4)) = ((y^2 + 16)/(y^2 - 16))
(((2y)(y - 4)) - ((2y)(y + 4)))/((y - 4)(y + 4)) = ((y^2 + 16)/(y^2 - 16))
(2y^2 - 8y - 2y^2 - 8y)/(y^2 - 16) = (y^2 + 16)/(y^2 - 16)
(-16y)/(y^2 - 16) = (y^2 + 16)/(y^2 - 16)
-16y = y^2 + 16
y^2 + 16y + 16 = 0

Except for the mistyping, you have it correct so far, but here is the rest of it

y = (-b ± sqrt(b^2 - 4ac))/2a

y = (-16 ± sqrt(16^2 - 4(1)(16)))/(2(1))
y = (-16 ± sqrt(256 - 64))/2
y = (-16 ± sqrt(192))/2
y = (-16 ± sqrt(64 * 3))/2
y = (-16 ± 8sqrt(3))//2
y = -8 ± 4sqrt(3)

ANS : y = -8 ± 4sqrt(3)

Answer 2

Assume you meant

2y/(y + 4) - 2y/(y - 4) = (y^2 + 16)/(y^2 - 16)

y^2 - 16, a difference of 2 squares, factors into (y + 4)(y - 4), which is a common denominator for all the fractions in the equation.

On your second line you appeared to have multiplied the left side by this common denominator, but you neglected to show you multiplied the right side also. My second line is

2y(y - 4) - 2y(y + 4) = y^2 + 16

From there

2y^2 - 8y - 2y^2 - 8y) = y^2 + 16
-16y = y^2 + 16
0 = y^2 + 16y + 16

or

y^2 + 16y + 16 = 0

Are you supposed to solve this quadratic equation, or just simplify it?

Answer 3

2y/y+4 -2y/y-4 = y^2+16/y^2-16

2y(y-4) – 2y(y+4) = y^2+16/(y+4)(y-4)

2y^2-8y-2y^2-8y = y^2+16

-16y = y^2+16

i dont know how to finish it but u need to divide the right side of the equal sign by the left side.
so it should be:
y = ???????

im not a math genius but i hope i can help

Answer 4

Your second equation is not correct. You are doing 2 things at the same time - try breaking down into 2 steps.

Your third equation is correct as well as the next equation.

Your last equation is not correct - you are missing something. Start from the 4th equation and carefully bring the term on the left to the righthand side.

Then you will have simplified the equation - now solve for y.

It seems you have the right idea - you just have to be careful how you execute. Keep the steps simple until you get the hang of it.

Besides this question has been asked before - but you should be able to solve it your own.

Answer 5

2y/y+4 -2y/y-4 = y^2+16/y^2-16

2y(y-4) – 2y(y+4) = y^2+16/(y+4)(y-4)

2y^2-8y-2y^2-8y = y^2+16

-16y = y^2+16

0 = y^2+16 + 16
........

So ??

Answer 6

There's a mistake (or a typo)

Shouldn't it be 0 = y^2 +16y + 16

Answer 7

You dropped a y in the end, it should be
0=y^2 + 16y + 16

Answer 8

should be y^2 + 16y +16 = 0, which you have to solve with the quadratic equation to find the solutions for y

Answer 9

Is it (16/y^2) -16
or
16/(y^2-16)
or
(y^2 + 16)/(y^2 -16)


You need to know the differences. I assume that's the third option. You are missing a y in the second term.

Answer 10

If the answers are incorrect, your teacher tomorrow will go over with you why. If we tell you, not that I know anything about algebra, what the correct answer is then you cannot learn why you came up with the wrong answer and how to get the correct answer. If you are correct then you know you did it right. Just do it again, and if you come up with the same answer then you can be almost 100% sure that it is correct. If it is wrong, come in with both answers and see what the teacher says.