Solve 3xy+y=12x

$$3 x y + y = 12 x$$

Solve for x

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

$$y\neq 4$$

Steps for Solving Linear Equation

Subtract 12x from both sides.

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

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

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

Combine all terms containing x.

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

Divide both sides by 3y-12.

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

Dividing by 3y-12 undoes the multiplication by 3y-12.

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

Divide -y by 3y-12.

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

Solve for y

$$y=\frac{12x}{3x+1}$$

$$x\neq -\frac{1}{3}$$

Steps for Solving Linear Equation

Combine all terms containing y.

$$\left(3x+1\right)y=12x$$

Divide both sides by 3x+1.

$$\frac{\left(3x+1\right)y}{3x+1}=\frac{12x}{3x+1}$$

Dividing by 3x+1 undoes the multiplication by 3x+1.

$$y=\frac{12x}{3x+1}$$