Solve z^2+(x+2)z+y(1-2)=0

$${ z }^{ 2 } +(x+2)z+y(1-2)=0$$

Solve for x

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

Steps for Solving Linear Equation

Use the distributive property to multiply x+2 by z.

$$z^{2}+xz+2z+y\left(1-2\right)=0$$

Subtract 2 from 1 to get -1.

$$z^{2}+xz+2z+y\left(-1\right)=0$$

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

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

Subtract 2z from both sides.

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

Subtract y\left(-1\right) from both sides.

$$xz=-z^{2}-2z-y\left(-1\right)$$

Multiply -1 and -1 to get 1.

$$xz=-z^{2}-2z+y$$

The equation is in standard form.

$$zx=y-z^{2}-2z$$

Divide both sides by z.

$$\frac{zx}{z}=\frac{y-z^{2}-2z}{z}$$

Dividing by z undoes the multiplication by z.

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

Divide -z^{2}-2z+y by z.

$$x=-z+\frac{y}{z}-2$$

Solve for y

$$y=z\left(x+z+2\right)$$

Steps for Solving Linear Equation

Use the distributive property to multiply x+2 by z.

$$z^{2}+xz+2z+y\left(1-2\right)=0$$

Subtract 2 from 1 to get -1.

$$z^{2}+xz+2z+y\left(-1\right)=0$$

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

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

Subtract xz from both sides.

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

Subtract 2z from both sides.

$$y\left(-1\right)=-z^{2}-xz-2z$$

The equation is in standard form.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

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

$$y=z\left(x+z+2\right)$$