Solve 2-y/y=(2-y)(x)/y^2+y

$$\frac { 2 - y } { y } = \frac { ( 2 - y ) ( x ) } { y ^ { 2 } + y }$$

Solve for x

$$\left\{\begin{matrix}x=y+1\text{, }&y\neq -1\text{ and }y\neq 0\\x\in \mathrm{R}\text{, }&y=2\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by y\left(y+1\right), the least common multiple of y,y^{2}+y.

$$\left(y+1\right)\left(2-y\right)=\left(2-y\right)x$$

Use the distributive property to multiply y+1 by 2-y and combine like terms.

$$y-y^{2}+2=\left(2-y\right)x$$

Use the distributive property to multiply 2-y by x.

$$y-y^{2}+2=2x-yx$$

Swap sides so that all variable terms are on the left hand side.

$$2x-yx=y-y^{2}+2$$

Combine all terms containing x.

$$\left(2-y\right)x=y-y^{2}+2$$

The equation is in standard form.

$$\left(2-y\right)x=2+y-y^{2}$$

Divide both sides by 2-y.

$$\frac{\left(2-y\right)x}{2-y}=-\frac{\left(y-2\right)\left(y+1\right)}{2-y}$$

Dividing by 2-y undoes the multiplication by 2-y.

$$x=-\frac{\left(y-2\right)\left(y+1\right)}{2-y}$$

Divide -\left(-2+y\right)\left(1+y\right) by 2-y.

$$x=y+1$$

Solve for y

$$\left\{\begin{matrix}\\y=2\text{, }&\text{unconditionally}\\y=x-1\text{, }&x\neq 0\text{ and }x\neq 1\end{matrix}\right.$$