Solve y^2-x=2x-1/y

$${ y }^{ 2 } -x=2x- \frac{ 1 }{ y }$$

Solve for x

$$x=\frac{y^{3}+1}{3y}$$

$$y\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by y.

$$yy^{2}-xy=2xy-1$$

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

$$y^{3}-xy=2xy-1$$

Subtract 2xy from both sides.

$$y^{3}-xy-2xy=-1$$

Combine -xy and -2xy to get -3xy.

$$y^{3}-3xy=-1$$

Subtract y^{3} from both sides.

$$-3xy=-1-y^{3}$$

The equation is in standard form.

$$\left(-3y\right)x=-y^{3}-1$$

Divide both sides by -3y.

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

Dividing by -3y undoes the multiplication by -3y.

$$x=\frac{-y^{3}-1}{-3y}$$

Divide -y^{3}-1 by -3y.

$$x=\frac{y^{2}}{3}+\frac{1}{3y}$$