Question

2-3y/3-y=y+2/y+3
Помогите пожалуйста

Answer 1

$\frac{2-3y}{3-y}=\frac{y+2}{y+3},~~y\neq3,~y\neq-3\\(2-3y)(y+3)=(y+2)(3-y)=0\\2y+6-3y^2-9y-y+y^2-6=0\\-8y-2y^2=0\\y(4+y)=0\\\\y_1= -4\\y_2=0$