Solve 5/3x-3=7yquad7y-8/3+5

$$\frac { 5 } { 3 } x - 3 = 7 y \quad 7 y - \frac { 8 } { 3 } + 5$$

Solve for y (complex solution)

$$\left\{\begin{matrix}y=-\frac{16-5x}{21y_{7}}\text{, }&y_{7}\neq 0\\y\in \mathrm{C}\text{, }&x=\frac{16}{5}\text{ and }y_{7}=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Add -\frac{8}{3} and 5 to get \frac{7}{3}.

$$\frac{5}{3}x-3=7y_{7}y+\frac{7}{3}$$

Swap sides so that all variable terms are on the left hand side.

$$7y_{7}y+\frac{7}{3}=\frac{5}{3}x-3$$

Subtract \frac{7}{3} from both sides.

$$7y_{7}y=\frac{5}{3}x-3-\frac{7}{3}$$

Subtract \frac{7}{3} from -3 to get -\frac{16}{3}.

$$7y_{7}y=\frac{5}{3}x-\frac{16}{3}$$

The equation is in standard form.

$$7y_{7}y=\frac{5x-16}{3}$$

Divide both sides by 7y_{7}.

$$\frac{7y_{7}y}{7y_{7}}=\frac{5x-16}{3\times 7y_{7}}$$

Dividing by 7y_{7} undoes the multiplication by 7y_{7}.

$$y=\frac{5x-16}{3\times 7y_{7}}$$

Divide \frac{-16+5x}{3} by 7y_{7}.

$$y=\frac{5x-16}{21y_{7}}$$

Solve for x

$$x=\frac{21yy_{7}+16}{5}$$

Steps for Solving Linear Equation

Add -\frac{8}{3} and 5 to get \frac{7}{3}.

$$\frac{5}{3}x-3=7y_{7}y+\frac{7}{3}$$

Add 3 to both sides.

$$\frac{5}{3}x=7y_{7}y+\frac{7}{3}+3$$

Add \frac{7}{3} and 3 to get \frac{16}{3}.

$$\frac{5}{3}x=7y_{7}y+\frac{16}{3}$$

The equation is in standard form.

$$\frac{5}{3}x=7yy_{7}+\frac{16}{3}$$

Divide both sides of the equation by \frac{5}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{5}{3}x}{\frac{5}{3}}=\frac{7yy_{7}+\frac{16}{3}}{\frac{5}{3}}$$

Dividing by \frac{5}{3} undoes the multiplication by \frac{5}{3}.

$$x=\frac{7yy_{7}+\frac{16}{3}}{\frac{5}{3}}$$

Divide 7y_{7}y+\frac{16}{3} by \frac{5}{3} by multiplying 7y_{7}y+\frac{16}{3} by the reciprocal of \frac{5}{3}.

$$x=\frac{21yy_{7}+16}{5}$$

Solve for y

$$\left\{\begin{matrix}y=-\frac{16-5x}{21y_{7}}\text{, }&y_{7}\neq 0\\y\in \mathrm{R}\text{, }&x=\frac{16}{5}\text{ and }y_{7}=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Add -\frac{8}{3} and 5 to get \frac{7}{3}.

$$\frac{5}{3}x-3=7y_{7}y+\frac{7}{3}$$

Swap sides so that all variable terms are on the left hand side.

$$7y_{7}y+\frac{7}{3}=\frac{5}{3}x-3$$

Subtract \frac{7}{3} from both sides.

$$7y_{7}y=\frac{5}{3}x-3-\frac{7}{3}$$

Subtract \frac{7}{3} from -3 to get -\frac{16}{3}.

$$7y_{7}y=\frac{5}{3}x-\frac{16}{3}$$

The equation is in standard form.

$$7y_{7}y=\frac{5x-16}{3}$$

Divide both sides by 7y_{7}.

$$\frac{7y_{7}y}{7y_{7}}=\frac{5x-16}{3\times 7y_{7}}$$

Dividing by 7y_{7} undoes the multiplication by 7y_{7}.

$$y=\frac{5x-16}{3\times 7y_{7}}$$

Divide \frac{5x-16}{3} by 7y_{7}.

$$y=\frac{5x-16}{21y_{7}}$$