Question 129175
Solve for y:
{{{(5/(y-3))-(30/(y^2-9)) = 1}}} Factor the second denominator.
{{{(5/(y-3))-(30/((y-3)(y+3))) = 1}}} Multiply the first fraction by {{{(y+3)/(y+3)}}} to get the fractions over a common denominator.
{{{(5(y+3)-30)/(y-3)(y+3) = 1}}} Multiply both sides by {{{(y-3)(y+3)}}}
{{{5(y+3)-30 = (y-3)(y+3)}}} Perform the indicated multiplication on both sides.
{{{5y+15-30 = y^2-9}}} Simplify.
{{{5y-15 = y^2-9}}} Subtract 5y from both sides.
{{{-15 = y^2-5y-9}}} Add 15 to both sides.
{{{y^2-5y+6 = 0}}} Factor this quadratic equation.
{{{(y-2)(y-3) = 0}}} Apply the zero product principle.
{{{y-2 = 0}}} or {{{y-3 = 0}}}, therefore...
{{{y = 2}}} or {{{y = 3}}}
 CAUTION!!!
The original equation is:
{{{5/(y-3) - 30/(y^2-9) = 1}}}
If you substitute y = 3, you will get:

{{{5/(3-3)-30/(9-9)}}} = {{{(5/0)-(30/0)}}}...and, as you know, division by zero is undefined and thus not allowed.
So you have only one valid solution to this problem: y = 2