Solve y=1+xyz/3x-2

$$y = \frac { 1 + x y z } { 3 x - 2 }$$

Solve for x

$$\left\{\begin{matrix}x=\frac{2y+1}{y\left(3-z\right)}\text{, }&y\neq 0\text{ and }z\neq 3\text{ and }\left(z=0\text{ or }y\neq -\frac{3}{2z}\right)\\x\neq \frac{2}{3}\text{, }&y=-\frac{1}{2}\text{ and }z=3\end{matrix}\right.$$

Steps for Solving Linear Equation

Variable x cannot be equal to \frac{2}{3} since division by zero is not defined. Multiply both sides of the equation by 3x-2.

$$y\left(3x-2\right)=1+xyz$$

Use the distributive property to multiply y by 3x-2.

$$3yx-2y=1+xyz$$

Subtract xyz from both sides.

$$3yx-2y-xyz=1$$

Add 2y to both sides.

$$3yx-xyz=1+2y$$

Combine all terms containing x.

$$\left(3y-yz\right)x=1+2y$$

The equation is in standard form.

$$\left(3y-yz\right)x=2y+1$$

Divide both sides by -yz+3y.

$$\frac{\left(3y-yz\right)x}{3y-yz}=\frac{2y+1}{3y-yz}$$

Dividing by -yz+3y undoes the multiplication by -yz+3y.

$$x=\frac{2y+1}{3y-yz}$$

Divide 2y+1 by -yz+3y.

$$x=\frac{2y+1}{y\left(3-z\right)}$$

Variable x cannot be equal to \frac{2}{3}.

$$x=\frac{2y+1}{y\left(3-z\right)}\text{, }x\neq \frac{2}{3}$$

Solve for y

$$y=\frac{1}{-xz+3x-2}$$

$$\left(z=3\text{ or }x\neq \frac{2}{3-z}\right)\text{ and }x\neq \frac{2}{3}$$

Steps for Solving Linear Equation

Subtract \frac{1+xyz}{3x-2} from both sides.

$$y-\frac{1+xyz}{3x-2}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply y times \frac{3x-2}{3x-2}.

$$\frac{y\left(3x-2\right)}{3x-2}-\frac{1+xyz}{3x-2}=0$$

Since \frac{y\left(3x-2\right)}{3x-2} and \frac{1+xyz}{3x-2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{y\left(3x-2\right)-\left(1+xyz\right)}{3x-2}=0$$

Do the multiplications in y\left(3x-2\right)-\left(1+xyz\right).

$$\frac{3yx-2y-1-xyz}{3x-2}=0$$

Multiply both sides of the equation by 3x-2.

$$3yx-2y-1-xyz=0$$

Add 1 to both sides. Anything plus zero gives itself.

$$3yx-2y-xyz=1$$

Combine all terms containing y.

$$\left(3x-2-xz\right)y=1$$

The equation is in standard form.

$$\left(-xz+3x-2\right)y=1$$

Divide both sides by 3x-2-xz.

$$\frac{\left(-xz+3x-2\right)y}{-xz+3x-2}=\frac{1}{-xz+3x-2}$$

Dividing by 3x-2-xz undoes the multiplication by 3x-2-xz.

$$y=\frac{1}{-xz+3x-2}$$