Solve y^2-x=1quad(yxy^2

$$y ^ { 2 } - x = 1 \quad ( y x y ^ { 2 }$$

Solve for x

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

$$y\neq -1$$

Steps for Solving Linear Equation

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

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

Subtract 1y^{3}x from both sides.

$$y^{2}-x-1y^{3}x=0$$

Reorder the terms.

$$-x+y^{2}-xy^{3}=0$$

Subtract y^{2} from both sides. Anything subtracted from zero gives its negation.

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

Combine all terms containing x.

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

The equation is in standard form.

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

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

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

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

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

Divide -y^{2} by -1-y^{3}.

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