Solve x^2-3x/y=x-3/x^2+2x

$$\frac{ { x }^{ 2 } -3x }{ y } = \frac{ x-3 }{ { x }^{ 2 } +2x }$$

Solve for y

$$\left\{\begin{matrix}y=\left(x+2\right)x^{2}\text{, }&x\neq -2\text{ and }x\neq 0\\y\neq 0\text{, }&x=3\end{matrix}\right.$$

Steps for Solving Linear Equation

Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by xy\left(x+2\right), the least common multiple of y,x^{2}+2x.

$$x\left(x+2\right)\left(x^{2}-3x\right)=y\left(x-3\right)$$

Use the distributive property to multiply x by x+2.

$$\left(x^{2}+2x\right)\left(x^{2}-3x\right)=y\left(x-3\right)$$

Use the distributive property to multiply x^{2}+2x by x^{2}-3x and combine like terms.

$$x^{4}-x^{3}-6x^{2}=y\left(x-3\right)$$

Use the distributive property to multiply y by x-3.

$$x^{4}-x^{3}-6x^{2}=yx-3y$$

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

$$yx-3y=x^{4}-x^{3}-6x^{2}$$

Combine all terms containing y.

$$\left(x-3\right)y=x^{4}-x^{3}-6x^{2}$$

Divide both sides by -3+x.

$$\frac{\left(x-3\right)y}{x-3}=\frac{\left(x-3\right)\left(x+2\right)x^{2}}{x-3}$$

Dividing by -3+x undoes the multiplication by -3+x.

$$y=\frac{\left(x-3\right)\left(x+2\right)x^{2}}{x-3}$$

Divide \left(-3+x\right)\left(2+x\right)x^{2} by -3+x.

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

Variable y cannot be equal to 0.

$$y=\left(x+2\right)x^{2}\text{, }y\neq 0$$