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
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
2 answers
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.
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.