Réitigh 3+xi/1+2i=y+2i

$$\frac { 3 + x i } { 1 + 2 i } = y + 2 i$$

Réitigh do ξ.

$$\xi =\left(1+2i\right)y+\left(-7+2i\right)$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Roinn 3+\xi faoi 1+2i chun \frac{3}{1+2i}+\frac{\xi }{1+2i} a fháil.

$$\frac{3}{1+2i}+\frac{\xi }{1+2i}=y+2i$$

Iolraigh uimhreoir agus ainmneoir \frac{3}{1+2i} faoi chomhchuingeach coimpléascach an ainmneora, 1-2i.

$$\frac{3\left(1-2i\right)}{\left(1+2i\right)\left(1-2i\right)}+\frac{\xi }{1+2i}=y+2i$$

Déan iolrúcháin in \frac{3\left(1-2i\right)}{\left(1+2i\right)\left(1-2i\right)}.

$$\frac{3-6i}{5}+\frac{\xi }{1+2i}=y+2i$$

Roinn 3-6i faoi 5 chun \frac{3}{5}-\frac{6}{5}i a fháil.

$$\frac{3}{5}-\frac{6}{5}i+\frac{\xi }{1+2i}=y+2i$$

Bain \frac{3}{5}-\frac{6}{5}i ón dá thaobh.

$$\frac{\xi }{1+2i}=y+2i-\left(\frac{3}{5}-\frac{6}{5}i\right)$$

Méadaigh -1 agus \frac{3}{5}-\frac{6}{5}i chun -\frac{3}{5}+\frac{6}{5}i a fháil.

$$\frac{\xi }{1+2i}=y+2i+\left(-\frac{3}{5}+\frac{6}{5}i\right)$$

Déan suimiú in 2i+\left(-\frac{3}{5}+\frac{6}{5}i\right).

$$\frac{\xi }{1+2i}=y-\frac{3}{5}+\frac{16}{5}i$$

Tá an chothromóid i bhfoirm chaighdeánach.

$$\left(\frac{1}{5}-\frac{2}{5}i\right)\xi =y+\left(-\frac{3}{5}+\frac{16}{5}i\right)$$

Roinn an dá thaobh faoi \frac{1}{5}-\frac{2}{5}i.

$$\frac{\left(\frac{1}{5}-\frac{2}{5}i\right)\xi }{\frac{1}{5}-\frac{2}{5}i}=\frac{y+\left(-\frac{3}{5}+\frac{16}{5}i\right)}{\frac{1}{5}-\frac{2}{5}i}$$

Má roinntear é faoi \frac{1}{5}-\frac{2}{5}i cuirtear an iolrúchán faoi \frac{1}{5}-\frac{2}{5}i ar ceal.

$$\xi =\frac{y+\left(-\frac{3}{5}+\frac{16}{5}i\right)}{\frac{1}{5}-\frac{2}{5}i}$$

Roinn y+\left(-\frac{3}{5}+\frac{16}{5}i\right) faoi \frac{1}{5}-\frac{2}{5}i.

$$\xi =\left(1+2i\right)y+\left(-7+2i\right)$$

Réitigh do y.

$$y=\left(\frac{1}{5}-\frac{2}{5}i\right)\xi +\left(\frac{3}{5}-\frac{16}{5}i\right)$$