Question

y+1/y-1 - y-1/y+1 = 8/3

I know that I have to multiply by the LCD of the denominaters, but what is the LCD of y-1, y+1, and 3?????

PLEASE HELP!!
~aShLeY

Answer 1

You want the denominator to be the least common multiple, LCM.
That LCM is 3(y-1)(y+1).
Once the denominators are the same on both sides, you can write this equation for the numerators only:

3(y+1)^2 - 3(y-1)^2 = 8(y+1)(y-1)
12y = 8 y^2 - 8
2y^2 -3y -2 = 0
(2y +1)(y -2) = 0
y = 2 or -1/2

Answer 2

3(y+1)(y-1) = 3 (y^2-1)
but you do not really need to know tha, use it in the original form.

for example the first term:

3(y+1)(y-1) [ (y+1)/(y-1) ]
is
3 (y+1)(y+1) = 3 (y+1)^2

the second term will give similarly

- 3 (y-1)^1

the comibation is then (again the difference of two squares)as in (A^2-B^2)=(A+B)(A-B)

3 [ (y+1)^2 -(y-1)^2 ]

which is

3 [ (y+1)+(y-1)] [ (y+1)-(y-1) ]

= 3 [ (2y)(2)] = 12 y

I will leave the right side for you to do. Check my arithmetic.