Solve z=xy+x+y-2

$$z = x y + x + y - 2$$

Solve for x

$$\left\{\begin{matrix}x=-\frac{y-z-2}{y+1}\text{, }&y\neq -1\\x\in \mathrm{R}\text{, }&z=-3\text{ and }y=-1\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$xy+x+y-2=z$$

Subtract y from both sides.

$$xy+x-2=z-y$$

Add 2 to both sides.

$$xy+x=z-y+2$$

Combine all terms containing x.

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

The equation is in standard form.

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

Divide both sides by y+1.

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

Dividing by y+1 undoes the multiplication by y+1.

$$x=\frac{2+z-y}{y+1}$$

Solve for y

$$\left\{\begin{matrix}y=-\frac{x-z-2}{x+1}\text{, }&x\neq -1\\y\in \mathrm{R}\text{, }&z=-3\text{ and }x=-1\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$xy+x+y-2=z$$

Subtract x from both sides.

$$xy+y-2=z-x$$

Add 2 to both sides.

$$xy+y=z-x+2$$

Combine all terms containing y.

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

The equation is in standard form.

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

Divide both sides by x+1.

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

Dividing by x+1 undoes the multiplication by x+1.

$$y=\frac{2+z-x}{x+1}$$