how do you slove this equation ?

Question

2/(y-1) + 6/(y+1) = 4, where y is not equals to 1 or -1.

Answer 1

Since we know y is not equal to 1 or -1, we can just multiply everything by the lowest common denominator LCD. In this case, it would be (y - 1) (y + 1).

2/(y - 1) + 6/(y + 1) = 4

Multiplying both sides by (y - 1)(y + 1) will eliminate all fractions. We will, however, be left with stuff:

2(y + 1) + 6(y - 1) = 4(y - 1)(y + 1)

Expand, to get

2y + 2 + 6y - 6 = 4(y^2 - 1)

Simplifying,

8y - 4 = 4y^2 - 4

Moving everything to the right hand side, we get

0 = 4y^2 - 4 - 8y + 4
0 = 4y^2 - 8y

Dividing both sides by 4,

0 = y^2 - 2y

And factoring,

0 = y(y - 2)

Therefore, y = {0, 2}

Answer 2

2/(y-1) + 6/(y+1) = 4

Multiplying throughout by the LCM (y-1)*(y+1) we get

2(y+1) + 6(y-1) = 4(y+1)(y-1)

2y + 2 + 6y - 6 = 4y^2 - 4

4y^2 - 4 = 0

4y(y-2) = 0

So y = 0 OR y = 2

Answer 3

the common denominator is (y^2-1) so you have
(2y+2 +6y-6)(y^2-1)=4====> 8y-4=4y^2-4==== (y^2-2y)=0
y(y-2)=0 so y=0 . y=2

Answer 4

1/y-1 + 3/y+1 = 2
y+1+3y-3 = 2(y^2 - 1)
2y^2 - 4y = 0
y = 0 or y = 2

Answer 5

2/(y-1) +6 (y+1) = 4

2(y+1) +6(y-1)= 4(y-1)(y+1)

8y-4=4(y^2-1)
8y-4=4y^2-4
8y=4y^2
2y=y^2
y^2-2y=0
y(y-2)=0

either y=0 or y=2