Solve yx^2+zx+x=0

$$y { x }^{ 2 } +zx+x=0$$

Solve for y

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

Steps for Solving Linear Equation

Subtract zx from both sides. Anything subtracted from zero gives its negation.

$$yx^{2}+x=-zx$$

Subtract x from both sides.

$$yx^{2}=-zx-x$$

Reorder the terms.

$$yx^{2}=-xz-x$$

The equation is in standard form.

$$x^{2}y=-xz-x$$

Divide both sides by x^{2}.

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

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

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

Divide -x\left(1+z\right) by x^{2}.

$$y=-\frac{z+1}{x}$$

Solve for x

$$\left\{\begin{matrix}\\x=0\text{, }&\text{unconditionally}\\x=-\frac{z+1}{y}\text{, }&y\neq 0\\x\in \mathrm{R}\text{, }&z=-1\text{ and }y=0\end{matrix}\right.$$