Solve x/13+x/y=20

$$\frac { x } { 13 } + \frac { x } { y } = 20$$

Solve for x

$$x=\frac{260y}{y+13}$$

$$y\neq -13\text{ and }y\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 13y, the least common multiple of 13,y.

$$yx+13x=260y$$

Combine all terms containing x.

$$\left(y+13\right)x=260y$$

Divide both sides by y+13.

$$\frac{\left(y+13\right)x}{y+13}=\frac{260y}{y+13}$$

Dividing by y+13 undoes the multiplication by y+13.

$$x=\frac{260y}{y+13}$$

Solve for y

$$y=-\frac{13x}{x-260}$$

$$x\neq 0\text{ and }x\neq 260$$

Steps for Solving Linear Equation

Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 13y, the least common multiple of 13,y.

$$yx+13x=260y$$

Subtract 260y from both sides.

$$yx+13x-260y=0$$

Subtract 13x from both sides. Anything subtracted from zero gives its negation.

$$yx-260y=-13x$$

Combine all terms containing y.

$$\left(x-260\right)y=-13x$$

Divide both sides by x-260.

$$\frac{\left(x-260\right)y}{x-260}=-\frac{13x}{x-260}$$

Dividing by x-260 undoes the multiplication by x-260.

$$y=-\frac{13x}{x-260}$$

Variable y cannot be equal to 0.

$$y=-\frac{13x}{x-260}\text{, }y\neq 0$$