Question 329664
solve
{{{3/(y-3) - 7/(y+2) + 4 = 0 }}}
:
It's an equation so we can get rid of the denominators by mult by (y-3)(y+2)
(y-3)(y+2)*{{{3/(y-3)}}} - (y-3)(y+2)*{{{7/(y+2)}}} + 4(y-3)(y+2) = 0
:
Cancel the denominators, FOIL
3(y+2) -  7(y-3) + 4(y^2 - y - 6) = 0
:
3y + 6 - 7y + 21 + 4y^2 - 4y - 24 = 0
:
Combine like terms, form a quadratic equation
4y^2 + 3y - 7y - 4y + 6 + 21 - 24 = 0
:
4y^2 - 8y + 3 = 0
Factor
(2y - 3)(2y - 1)  = 0
Two solutions
2y = 3
y = 1.5
and
2y = 1
y = .5
:
:
Check solution using x=1.5 in the original equation
{{{3/(1.5-3) - 7/(1.5+2) + 4 = 0 }}}
{{{3/(-1.5) - 7/(3.5) + 4 = 0 }}}
-2 - 2 + 4 = 0
:
You can check the x=.5 solution