Решите уравнение: Помогите пожалуйста срочно нужно

$2). \frac{2-3y}{3-y}= \frac{y-2}{y+3}; y \neq 3; y \neq -3\\(2-3y)*(y+3)=(y+2)*(3-y)\\(2-3y)*(y+3)-(y+2)*(3-y)=0\\2y=6-3y^{2} -9y-(y-y^{2}+6)=0\\2y+6-3y^{2}-9y-y+y^{2}-6=0\\-8y-2y^{2}=0\\-2y*(4+y)=0\\y*(4+y)=0\\y=0;\\\\4+y=0\\y=0;\\y=-4;$

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$\frac{7x+5}{x-1}- \frac{2x}{x+1}= -5; x \neq 1; x\neq -1;\\\frac{7x+5}{x-1}- \frac{2x}{x+1}+5=0\\\frac{(x+1)*(7x+5)-2x(x-1)+5(x-1)(x+1)}{(x-1)(x+1)}=0\\\\\frac{10x^{2}+14x}{(x-1)(x+1)}=0\\ 10x^{2}=14x=0\\2x(5x+7)=0\\x1=0;\\5x+7=0\\x1=0\\x2=-\frac{7}{5}$