Solve 2xy-y^2=-xdivy^2+2divy

$$2xy- { y }^{ 2 } =-x \div { y }^{ 2 } +2 \div y$$

Solve for x

$$x=\frac{y\left(y^{3}+2\right)}{2y^{3}+1}$$

$$y\neq -\frac{2^{\frac{2}{3}}}{2}\text{ and }y\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by y^{2}, the least common multiple of y^{2},y.

$$2xyy^{2}-y^{2}y^{2}=-x+y\times 2$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

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

To multiply powers of the same base, add their exponents. Add 2 and 2 to get 4.

$$2xy^{3}-y^{4}=-x+y\times 2$$

Add x to both sides.

$$2xy^{3}-y^{4}+x=y\times 2$$

Add y^{4} to both sides.

$$2xy^{3}+x=y\times 2+y^{4}$$

Combine all terms containing x.

$$\left(2y^{3}+1\right)x=y\times 2+y^{4}$$

The equation is in standard form.

$$\left(2y^{3}+1\right)x=y^{4}+2y$$

Divide both sides by 2y^{3}+1.

$$\frac{\left(2y^{3}+1\right)x}{2y^{3}+1}=\frac{y\left(y^{3}+2\right)}{2y^{3}+1}$$

Dividing by 2y^{3}+1 undoes the multiplication by 2y^{3}+1.

$$x=\frac{y\left(y^{3}+2\right)}{2y^{3}+1}$$