Solve x/5+36/y=500

$$\frac { x } { 5 } + \frac { 36 } { y } = 500$$

Solve for x

$$x=2500-\frac{180}{y}$$

$$y\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 5y, the least common multiple of 5,y.

$$yx+5\times 36=2500y$$

Multiply 5 and 36 to get 180.

$$yx+180=2500y$$

Subtract 180 from both sides.

$$yx=2500y-180$$

Divide both sides by y.

$$\frac{yx}{y}=\frac{2500y-180}{y}$$

Dividing by y undoes the multiplication by y.

$$x=\frac{2500y-180}{y}$$

Divide 2500y-180 by y.

$$x=2500-\frac{180}{y}$$

Solve for y

$$y=-\frac{180}{x-2500}$$

$$x\neq 2500$$

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 5y, the least common multiple of 5,y.

$$yx+5\times 36=2500y$$

Multiply 5 and 36 to get 180.

$$yx+180=2500y$$

Subtract 2500y from both sides.

$$yx+180-2500y=0$$

Subtract 180 from both sides. Anything subtracted from zero gives its negation.

$$yx-2500y=-180$$

Combine all terms containing y.

$$\left(x-2500\right)y=-180$$

Divide both sides by -2500+x.

$$\frac{\left(x-2500\right)y}{x-2500}=-\frac{180}{x-2500}$$

Dividing by -2500+x undoes the multiplication by -2500+x.

$$y=-\frac{180}{x-2500}$$

Variable y cannot be equal to 0.

$$y=-\frac{180}{x-2500}\text{, }y\neq 0$$