Question

$4y+3/y-7=y^{2}/7-y$

Answer 1

$frac{4y+3}{y-7}=frac{y^2}{7-y}\frac{4y+3}{y-7}+frac{y^2}{y-7}=0\frac{4y+3+y^2}{y-7}=0\y^2+4y+3=0\D=b^2-4ac=16-4*3=4\y_{_1}=-3\y_{_2}=-1$