Solve 3/y+x=4

$$\frac { 3 } { y } + x = 4$$

Solve for x

$$x=4-\frac{3}{y}$$

$$y\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by y.

$$3+yx=4y$$

Subtract 3 from both sides.

$$yx=4y-3$$

Divide both sides by y.

$$\frac{yx}{y}=\frac{4y-3}{y}$$

Dividing by y undoes the multiplication by y.

$$x=\frac{4y-3}{y}$$

Divide 4y-3 by y.

$$x=4-\frac{3}{y}$$

Solve for y

$$y=-\frac{3}{x-4}$$

$$x\neq 4$$

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 y.

$$3+yx=4y$$

Subtract 4y from both sides.

$$3+yx-4y=0$$

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

$$yx-4y=-3$$

Combine all terms containing y.

$$\left(x-4\right)y=-3$$

Divide both sides by x-4.

$$\frac{\left(x-4\right)y}{x-4}=-\frac{3}{x-4}$$

Dividing by x-4 undoes the multiplication by x-4.

$$y=-\frac{3}{x-4}$$

Variable y cannot be equal to 0.

$$y=-\frac{3}{x-4}\text{, }y\neq 0$$