Algebra 2 Question??

Question

(y/y-2) - (2/y+3)=10/(y+3)(y-2)

Answer 1

(y/y-2) - (2/y+3) = 10/(y+3)(y-2)
=> (y(y+3) - 2(y-2))/(y-2)(y+3) = 10/(y+3)(y-2)
=> y(y+3) - 2(y-2) = 10 as the denominator cancels out
=> y^2 + 3y -2y + 4 = 10
=> y^2 + y + 4 - 10 =0
=> y^2 + y - 6 = 0
=> y^2 + 3y -2y - 6 = 0
=> (y+3)(y-2) = 0

Therefore, y = -3 or y = 2

Answer 2

I think your bracketing is misleading: I suspect the question asks you to solve

y/(y-2) - 2/(y+3) = 10/[(y+3)(y-2)]

The simplest way is to multiply through by (y+3) (y+2) at once, but I'll do it slightly differently by first putting everything on the LHS over a common denominator of (y-2)(y+3). This means we have to multiply the first term by (y+3)/(y+3) and the second term by (y-2)/(y-2):

[y(y+3)]/[(y-2)(y+3)] - [2(y-2)]/[(y+3)(y-2)] = 10/[(y+3)(y-2)]
<=> [y(y+3) - 2(y-2)] / [(y+3)(y-2)] = 10 / [(y+3)(y-2)]
<=> y(y+3) - 2(y-2) = 10, for y /= -3 or 2
<=> y^2 + 3y - 2y + 4 = 10
<=> y^2 + y - 6 = 0
<=> (y + 3)(y - 2) = 0
<=> y = -3 or 2.
But y cannot be -3 or 2, as then the original LHS and RHS are both undefined. So there are NO solutions.