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

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

Réitigh do p_1.

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

Céimeanna chun Cothromóidí Líneacha a Réiteach

Úsáid an t-airí dáileach chun v_{12}\left(-i\right) a mhéadú faoi ϕ_{12}+p_{1}-p_{2}.

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

Athraigh na taobhanna ionas go mbeidh na téarmaí inathraitheacha ar fad ar an taobh clé.

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

Bain -iv_{12}ϕ_{12} ón dá thaobh.

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

Bain iv_{12}p_{2} ón dá thaobh.

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

Méadaigh -1 agus -i chun i a fháil.

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

Tá an chothromóid i bhfoirm chaighdeánach.

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

Roinn an dá thaobh faoi -iv_{12}.

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

Má roinntear é faoi -iv_{12} cuirtear an iolrúchán faoi -iv_{12} ar ceal.

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

Roinn V_{12}+iv_{12}ϕ_{12}-iv_{12}p_{2} faoi -iv_{12}.

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

Réitigh do V_12.

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