Solve zx+2/3y=5

$$zx+ \frac{ 2 }{ 3 } y=5$$

Solve for x

$$\left\{\begin{matrix}x=-\frac{2y-15}{3z}\text{, }&z\neq 0\\x\in \mathrm{R}\text{, }&y=\frac{15}{2}\text{ and }z=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract \frac{2}{3}y from both sides.

$$zx=5-\frac{2}{3}y$$

The equation is in standard form.

$$zx=-\frac{2y}{3}+5$$

Divide both sides by z.

$$\frac{zx}{z}=\frac{-\frac{2y}{3}+5}{z}$$

Dividing by z undoes the multiplication by z.

$$x=\frac{-\frac{2y}{3}+5}{z}$$

Divide 5-\frac{2y}{3} by z.

$$x=\frac{15-2y}{3z}$$

Solve for y

$$y=\frac{15-3xz}{2}$$

Steps for Solving Linear Equation

Subtract zx from both sides.

$$\frac{2}{3}y=5-zx$$

The equation is in standard form.

$$\frac{2}{3}y=5-xz$$

Divide both sides of the equation by \frac{2}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{2}{3}y}{\frac{2}{3}}=\frac{5-xz}{\frac{2}{3}}$$

Dividing by \frac{2}{3} undoes the multiplication by \frac{2}{3}.

$$y=\frac{5-xz}{\frac{2}{3}}$$

Divide 5-zx by \frac{2}{3} by multiplying 5-zx by the reciprocal of \frac{2}{3}.

$$y=\frac{15-3xz}{2}$$