Question 178231
13) Solve 
{{{2/(x-3)}}} = {{{4/(x+3)}}} + {{{8/(x^2-9)}}}
Note that the third denominator is the "difference of squares"; factors to:
{{{2/(x-3)}}} = {{{4/(x+3)}}} + {{{8/((x-3)(x+3))}}}
this is includes the other two denominators multiply equation by (x-3)(x+3)
2(x+3) = 4(x-3) + 8
:
2x + 6 = 4x - 12 + 8
:
2x + 6 = 4x - 4
:
2x - 4x = -4 - 6
:
-2x = -10
:
x = +5 
:
always check solutions by substituting for the variable in the original equation
:
:
14) Solve
{{{13/(y-1)}}} – 3 = {{{1/(y-1)}}}
Multiply equation by (y-1), results:
13 - 3(y-1) = 1
:
13 - 3y + 3 = 1
;
-3y + 13 = 1
:
-3y = 1 - 13
:
-3y = -12
:
y = +4
:
:
15) Solve
{{{(x+3)/(x-1)}}} = {{{(x+2)/(x-3)}}} 
Cross multiply and you have:
(x+3)*(x-3) = (x-1)*(x+2)
FOIL
x^2 - 9 = x^2 + x - 2
:
x^2 - x^2 - x = -2 + 9
:
-x = +7
which is
x = -7
:
: Be sure to check all these solutions in the original equation