Solve {1/x-1/y=1/3

$$\{ \frac { 1 } { x } - \frac { 1 } { y } = \frac { 1 } { 3 }$$

Solve for x

$$x=\frac{3y}{y+3}$$

$$y\neq 0\text{ and }y\neq -3$$

Steps for Solving Linear Equation

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 3xy, the least common multiple of x,y,3.

$$3y-3x=xy$$

Subtract xy from both sides.

$$3y-3x-xy=0$$

Subtract 3y from both sides. Anything subtracted from zero gives its negation.

$$-3x-xy=-3y$$

Combine all terms containing x.

$$\left(-3-y\right)x=-3y$$

The equation is in standard form.

$$\left(-y-3\right)x=-3y$$

Divide both sides by -3-y.

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

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

$$x=-\frac{3y}{-y-3}$$

Divide -3y by -3-y.

$$x=\frac{3y}{y+3}$$

Variable x cannot be equal to 0.

$$x=\frac{3y}{y+3}\text{, }x\neq 0$$