Solve y=3x^2-x^3-x/x4

$$y = 3 { x }^{ 2 } - { x }^{ 3 } - \frac{ x }{ x4 }$$

Solve for x_4

$$\left\{\begin{matrix}x_{4}=-\frac{x}{x^{3}-3x^{2}+y}\text{, }&x\neq 0\text{ and }y\neq -\left(x-3\right)x^{2}\\x_{4}\neq 0\text{, }&y=0\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Variable x_{4} cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x_{4}.

$$yx_{4}=3x^{2}x_{4}-x^{3}x_{4}-x$$

Subtract 3x^{2}x_{4} from both sides.

$$yx_{4}-3x^{2}x_{4}=-x^{3}x_{4}-x$$

Add x^{3}x_{4} to both sides.

$$yx_{4}-3x^{2}x_{4}+x^{3}x_{4}=-x$$

Combine all terms containing x_{4}.

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

The equation is in standard form.

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

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

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

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

$$x_{4}=-\frac{x}{x^{3}-3x^{2}+y}$$

Variable x_{4} cannot be equal to 0.

$$x_{4}=-\frac{x}{x^{3}-3x^{2}+y}\text{, }x_{4}\neq 0$$