Solve zy=3x-2/2x+1

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

Solve for x

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

$$y=0\text{ or }z\neq \frac{3}{2y}$$

Steps for Solving Linear Equation

Variable x cannot be equal to -\frac{1}{2} since division by zero is not defined. Multiply both sides of the equation by 2x+1.

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

Use the distributive property to multiply zy by 2x+1.

$$2zyx+zy=3x-2$$

Subtract 3x from both sides.

$$2zyx+zy-3x=-2$$

Subtract zy from both sides.

$$2zyx-3x=-2-zy$$

Combine all terms containing x.

$$\left(2zy-3\right)x=-2-zy$$

The equation is in standard form.

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

Divide both sides by 2yz-3.

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

Dividing by 2yz-3 undoes the multiplication by 2yz-3.

$$x=\frac{-yz-2}{2yz-3}$$

Divide -2-zy by 2yz-3.

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

Variable x cannot be equal to -\frac{1}{2}.

$$x=-\frac{yz+2}{2yz-3}\text{, }x\neq -\frac{1}{2}$$

Solve for y

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

Steps for Solving Linear Equation

Multiply both sides of the equation by 2x+1.

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

Use the distributive property to multiply zy by 2x+1.

$$2zyx+zy=3x-2$$

Combine all terms containing y.

$$\left(2zx+z\right)y=3x-2$$

The equation is in standard form.

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

Divide both sides by 2zx+z.

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

Dividing by 2zx+z undoes the multiplication by 2zx+z.

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

Divide 3x-2 by 2zx+z.

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