Question

2у-2 + у+3 =5

у+3      у-3

Answer 1

$frac{(2y-2)(y-3)}{(y+3)(y-3)} + frac{(y+3)^2}{(y-3)(y+3)} = frac{2y^2-6y-2y+6+y^2+6y+9}{(y-3)(y+3)} =frac{3y^2-2y+15}{(y-3)(y+3)}=5$
$frac{3y^2-2y+15}{(y-3)(y+3)}=5$
$frac{3y^2-2y+15}{(y-3)(y+3)}- frac{5y^2-45}{(y-3)(y+3)} =0$
$frac{3y^2-2y+15}{(y-3)(y+3)}- frac{5y^2-45}{(y-3)(y+3)} = frac{-2y^2-2y+60}{(y-3)(y+3)} =0$
$2y^2+2y-60=0$
x1=5,x2=-6