Solve V_{12}=v_{12}(-i(phi_{12}+p_{1}-p_{2}))

$$V _ { 12 } = v _ { 12 } ( - i ( \phi _ { 12 } + p _ { 1 } - p _ { 2 } ) )$$

Solve for p_1

$$\left\{\begin{matrix}p_{1}=p_{2}-ϕ_{12}+\frac{iV_{12}}{v_{12}}\text{, }&v_{12}\neq 0\\p_{1}\in \mathrm{C}\text{, }&V_{12}=0\text{ and }v_{12}=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply v_{12}\left(-i\right) by ϕ_{12}+p_{1}-p_{2}.

$$V_{12}=-iv_{12}ϕ_{12}-iv_{12}p_{1}+iv_{12}p_{2}$$

Swap sides so that all variable terms are on the left hand side.

$$-iv_{12}ϕ_{12}-iv_{12}p_{1}+iv_{12}p_{2}=V_{12}$$

Subtract -iv_{12}ϕ_{12} from both sides.

$$-iv_{12}p_{1}+iv_{12}p_{2}=V_{12}-\left(-iv_{12}ϕ_{12}\right)$$

Subtract iv_{12}p_{2} from both sides.

$$-iv_{12}p_{1}=V_{12}-\left(-iv_{12}ϕ_{12}\right)-iv_{12}p_{2}$$

Multiply -1 and -i to get i.

$$-iv_{12}p_{1}=V_{12}+iv_{12}ϕ_{12}-iv_{12}p_{2}$$

The equation is in standard form.

$$\left(-iv_{12}\right)p_{1}=V_{12}+iv_{12}ϕ_{12}-ip_{2}v_{12}$$

Divide both sides by -iv_{12}.

$$\frac{\left(-iv_{12}\right)p_{1}}{-iv_{12}}=\frac{V_{12}+iv_{12}ϕ_{12}-ip_{2}v_{12}}{-iv_{12}}$$

Dividing by -iv_{12} undoes the multiplication by -iv_{12}.

$$p_{1}=\frac{V_{12}+iv_{12}ϕ_{12}-ip_{2}v_{12}}{-iv_{12}}$$

Divide V_{12}+iv_{12}ϕ_{12}-iv_{12}p_{2} by -iv_{12}.

$$p_{1}=p_{2}-ϕ_{12}+\frac{iV_{12}}{v_{12}}$$

Solve for V_12

$$V_{12}=-iv_{12}\left(p_{1}-p_{2}+ϕ_{12}\right)$$