i need some help solving these problemsthis is my first time on this site, I am in dire need of some help. I am over 50 and have to take this math to complete my degree....Please help help-
3x-6
________
x^2 - 4x +4
3/y-3 - 4/y+3
$\displaystyle \frac{3}{y-3}-\frac{4}{y+3}$
$\displaystyle =\frac{3(y+3)}{(y-3)(y+3)}-\frac{4(y-3)}{(y+3)(y-3)}$
$\displaystyle =\frac{3(y+3)-4(y-3)}{(y-3)(y+3)}$
$\displaystyle =\frac{3y+9-4y+12}{(y-3)(y+3)}$
$\displaystyle =\frac{3-y}{(y-3)(y+3)}$
$\displaystyle =\frac{-(y-3)}{(y-3)(y+3)}$
$\displaystyle =\frac{-1}{y+3}$
Did you get it now ???