Question

помогите пожалуйста срочно

Answer 1

$\frac{2-33y}{y-4}=7y\\\\y-4\neq0\\\\y\neq4\\\\2-33y=7y*(y-4)\\\\2-33y=7y^{2}-28y\\\\7y^{2}+5y-2=0\\\\D=5^{2}-4*7*(-2)=25+56=81=9^{2}\\\\y_{1} =\frac{-5+9}{4}=\frac{2}{7}\\\\y_{2}=\frac{-5-9}{14}=-1$