Solve xy-4x-3y+12=0

$$xy-4x-3y+12 = 0$$

Solve for x (complex solution)

$$\left\{\begin{matrix}\\x=3\text{, }&\text{unconditionally}\\x\in \mathrm{C}\text{, }&y=4\end{matrix}\right.$$

Steps for Solving Linear Equation

Add 3y to both sides. Anything plus zero gives itself.

$$xy-4x+12=3y$$

Subtract 12 from both sides.

$$xy-4x=3y-12$$

Combine all terms containing x.

$$\left(y-4\right)x=3y-12$$

Divide both sides by -4+y.

$$\frac{\left(y-4\right)x}{y-4}=\frac{3y-12}{y-4}$$

Dividing by -4+y undoes the multiplication by -4+y.

$$x=\frac{3y-12}{y-4}$$

Divide -12+3y by -4+y.

$$x=3$$

Solve for y (complex solution)

$$\left\{\begin{matrix}\\y=4\text{, }&\text{unconditionally}\\y\in \mathrm{C}\text{, }&x=3\end{matrix}\right.$$

Steps for Solving Linear Equation

Add 4x to both sides. Anything plus zero gives itself.

$$xy-3y+12=4x$$

Subtract 12 from both sides.

$$xy-3y=4x-12$$

Combine all terms containing y.

$$\left(x-3\right)y=4x-12$$

Divide both sides by -3+x.

$$\frac{\left(x-3\right)y}{x-3}=\frac{4x-12}{x-3}$$

Dividing by -3+x undoes the multiplication by -3+x.

$$y=\frac{4x-12}{x-3}$$

Divide -12+4x by -3+x.

$$y=4$$

Solve for x

$$\left\{\begin{matrix}\\x=3\text{, }&\text{unconditionally}\\x\in \mathrm{R}\text{, }&y=4\end{matrix}\right.$$

Steps for Solving Linear Equation

Add 3y to both sides. Anything plus zero gives itself.

$$xy-4x+12=3y$$

Subtract 12 from both sides.

$$xy-4x=3y-12$$

Combine all terms containing x.

$$\left(y-4\right)x=3y-12$$

Divide both sides by -4+y.

$$\frac{\left(y-4\right)x}{y-4}=\frac{3y-12}{y-4}$$

Dividing by -4+y undoes the multiplication by -4+y.

$$x=\frac{3y-12}{y-4}$$

Divide -12+3y by -4+y.

$$x=3$$

Solve for y

$$\left\{\begin{matrix}\\y=4\text{, }&\text{unconditionally}\\y\in \mathrm{R}\text{, }&x=3\end{matrix}\right.$$

Steps for Solving Linear Equation

Add 4x to both sides. Anything plus zero gives itself.

$$xy-3y+12=4x$$

Subtract 12 from both sides.

$$xy-3y=4x-12$$

Combine all terms containing y.

$$\left(x-3\right)y=4x-12$$

Divide both sides by -3+x.

$$\frac{\left(x-3\right)y}{x-3}=\frac{4x-12}{x-3}$$

Dividing by -3+x undoes the multiplication by -3+x.

$$y=\frac{4x-12}{x-3}$$

Divide -12+4x by -3+x.

$$y=4$$