Solve (4x-y):(2x+3y)=4x-y/2x+3y=4x-y/y/frac{2x+3y{y}}

$$( 4 x - y ) : ( 2 x + 3 y ) = \frac { 4 x - y } { 2 x + 3 y } = \frac { \frac { 4 x - y } { y } } { \frac { 2 x + 3 y } { y } }$$

Evaluate (complex solution)

Solution Steps

Divide \frac{4x-y}{y} by \frac{2x+3y}{y} by multiplying \frac{4x-y}{y} by the reciprocal of \frac{2x+3y}{y}.

$$\frac{4x-y}{2x+3y}=\frac{4x-y}{2x+3y}\text{ and }\frac{4x-y}{2x+3y}=\frac{\left(4x-y\right)y}{y\left(2x+3y\right)}$$

Cancel out y in both numerator and denominator.

$$\frac{4x-y}{2x+3y}=\frac{4x-y}{2x+3y}\text{ and }\frac{4x-y}{2x+3y}=\frac{4x-y}{2x+3y}$$

Subtract \frac{4x-y}{2x+3y} from both sides.

$$\frac{4x-y}{2x+3y}-\frac{4x-y}{2x+3y}=0\text{ and }\frac{4x-y}{2x+3y}=\frac{4x-y}{2x+3y}$$

Subtract \frac{4x-y}{2x+3y} from \frac{4x-y}{2x+3y} to get 0.

$$0=0\text{ and }\frac{4x-y}{2x+3y}=\frac{4x-y}{2x+3y}$$

Compare 0 and 0.

$$\text{true}\text{ and }\frac{4x-y}{2x+3y}=\frac{4x-y}{2x+3y}$$

Subtract \frac{4x-y}{2x+3y} from both sides.

$$\text{true}\text{ and }\frac{4x-y}{2x+3y}-\frac{4x-y}{2x+3y}=0$$

Subtract \frac{4x-y}{2x+3y} from \frac{4x-y}{2x+3y} to get 0.

$$\text{true}\text{ and }0=0$$

Compare 0 and 0.

$$\text{true}\text{ and }\text{true}$$

The conjunction of \text{true} and \text{true} is \text{true}.

$$\text{true}$$

Solve for x

$$x\neq -\frac{3y}{2}$$

$$y\neq 0$$

Solve for y

$$y\in \mathrm{R}\setminus -\frac{2x}{3},0$$