Solve x^5-2x^4(y)x^3+2x^2=0

$$x ^ { 5 } - 2 x ^ { 4 } ( y ) x ^ { 3 } + 2 x ^ { 2 } = 0$$

Solve for y

$$\left\{\begin{matrix}y=\frac{1}{2x^{2}}+\frac{1}{x^{5}}\text{, }&x\neq 0\\y\in \mathrm{R}\text{, }&x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

To multiply powers of the same base, add their exponents. Add 4 and 3 to get 7.

$$x^{5}-2x^{7}y+2x^{2}=0$$

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

$$x^{5}-2x^{7}y=-2x^{2}$$

Subtract x^{5} from both sides.

$$-2x^{7}y=-2x^{2}-x^{5}$$

The equation is in standard form.

$$\left(-2x^{7}\right)y=-x^{5}-2x^{2}$$

Divide both sides by -2x^{7}.

$$\frac{\left(-2x^{7}\right)y}{-2x^{7}}=-\frac{x^{2}\left(x^{3}+2\right)}{-2x^{7}}$$

Dividing by -2x^{7} undoes the multiplication by -2x^{7}.

$$y=-\frac{x^{2}\left(x^{3}+2\right)}{-2x^{7}}$$

Divide -\left(2+x^{3}\right)x^{2} by -2x^{7}.

$$y=\frac{1}{2x^{2}}+\frac{1}{x^{5}}$$