Solve 3d/frac{d{v_{1}}+d/v_{2}+d/v_{2}}=frac{3v_{1}v_{2}}{2v_{1}+v_{2}}

$$\frac { 3 d } { \frac { d } { v _ { 1 } } + \frac { d } { v _ { 2 } } + \frac { d } { v _ { 2 } } } = \frac { 3 v _ { 1 } v _ { 2 } } { 2 v _ { 1 } + v _ { 2 } }$$

Solve for d

$$d\neq 0$$

$$v_{1}\neq -\frac{v_{2}}{2}\text{ and }v_{1}\neq 0\text{ and }v_{2}\neq 0$$

Solution Steps

Multiply both sides of the equation by 2v_{1}+v_{2}.

$$d^{-1}v_{1}v_{2}\times 3d=3v_{1}v_{2}$$

Reorder the terms.

$$3\times \frac{1}{d}dv_{1}v_{2}=3v_{1}v_{2}$$

Cancel out 3 on both sides.

$$\frac{1}{d}dv_{1}v_{2}=v_{1}v_{2}$$

Variable d cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by d.

$$1dv_{1}v_{2}=v_{1}v_{2}d$$

Subtract v_{1}v_{2}d from both sides.

$$1dv_{1}v_{2}-v_{1}v_{2}d=0$$

Combine 1dv_{1}v_{2} and -v_{1}v_{2}d to get 0.

$$0=0$$

Compare 0 and 0.

$$\text{true}$$

This is true for any d.

$$d\in \mathrm{R}$$

Variable d cannot be equal to 0.

$$d\in \mathrm{R}\setminus 0$$

Solve for v_1 (complex solution)

$$v_{1}\in \mathrm{C}\setminus -\frac{v_{2}}{2},0$$

$$d\neq 0\text{ and }v_{2}\neq 0$$

Solve for v_1

$$v_{1}\in \mathrm{R}\setminus -\frac{v_{2}}{2},0$$

$$d\neq 0\text{ and }v_{2}\neq 0$$