Answer
y= 12, -1
Work Step by Step
$\frac{4y}{y-3}-3=\frac{3y-1}{y+3}$
$(y-3)(y+3)*(\frac{4y}{y-3}-3=\frac{3y-1}{y+3})$
$4y(y+3)-3(y-3)(y+3)=(3y-1)(y-3)$
$4y^2+12y-3(y^2-9)=3y^2-9y-y+3$
$4y^2+12y-3y^2+27=3y^2-10y+3$
$y^2+12y+27=3y^2-10y+3$
$2y^2-22y-24=0$
$2(y^2-11y-12=0)$
$y^2-11y-12=0$
$(y-12)(y+1)=0$
$y-12=0$
$y=12$
$y+1=0$
$y=-1$
$y=12$
$\frac{4y}{y-3}-3=\frac{3y-1}{y+3}$
$\frac{4*12}{12-3}-3=\frac{3*12-1}{12+3}$
$48/9 - 3 = 35/15$
$16/3 - 3 = 7/3$
$16/3 - 9/3 = 7/3$
$y=-1$
$\frac{4y}{y-3}-3=\frac{3y-1}{y+3}$
$\frac{4*-1}{-1-3}-3=\frac{3*-1-1}{-1+3}$
$-4/-4 -3 = -4/2$
$1-3 =-2$
