How do u solve this equation?

Question

2/y-2+1/y+2=4/y2-4

whats the answer and how do u do it.

Answer 1

Multiply each fraction by a number to get a common denominator. In this case your common denominator should be y^2-4.

Then you can cancel out all of the denominators once they are the same and solve the problem.

2(y+2)+1(y-2)=4
2y+4+y-2=4
3y = 2
y = 2/3

Answer 2

Using brackets to clarify question:-
2 / (y - 2) + 1 / (y + 2) = 4 / (y - 2) (y + 2)
2 (y + 2) + 1 ( y - 2) = 4
2y + 4 + y - 2 = 4
3y = 2
y = 2 / 3

Answer 3

I think what you mean is:
2/(y-2) + 1/(y+2) = 4/(y^2-4)

Then multiply both sides by (y^2-4),
and notice that (y^2 - 4) = (y + 2)*(y - 2):
2(y+2) + (y-2) = 4
2y + 4 + y - 2 = 4
3y + 2 = 4
3y = 2
y = 2/3

Answer 4

common denominator first

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

then clear fractions

2y+4 + y-2 = 4

3y + 2 = 4

3y = 2

y = 2/3

Answer 5

2/y-2 +1/y+2 =4/y2-4
it means
2/y-2 +1/y+2 =4/2y-4
{2(y+2)+1(y-2)}/y^-4 =4/2y-4
y+2(2y-4)=4(y^-4)
y=+2or y= -2.

Answer 6

2/(y-2)+1/(y+2)=4/(y^2-4)
so {2(y+2)+1(y-2)}/(y-2)(y+2)=4/(y^2-4)
or{ 2y+4+y-2}/(y^2-4)=4/(y^2-4)-----using (a-b)(a+b)=a^2-b^2
or 3y+2=4-------cancellingy^2-4 on both sides
or 3y=2
or y=2/3 ans

Answer 7

2/y-2+1/y+2=4/y^2-4--------*(y+2)(y-2)
2(y+2)+1(y-2)=4
2y+4+y-2=4
3y+2=4
3y=4-2
3y=2
y=2/3