Réitigh (2x+i)(4+3i)=(5+yi)

$$( 2 x + i ) ( 4 + 3 i ) = ( 5 + y i )$$

Réitigh do x.

$$x=\left(\frac{3}{50}+\frac{2}{25}i\right)y+\left(\frac{2}{5}-\frac{4}{5}i\right)$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Úsáid an t-airí dáileach chun 2x+i a mhéadú faoi 4+3i.

$$\left(8+6i\right)x+\left(-3+4i\right)=5+yi$$

Bain -3+4i ón dá thaobh.

$$\left(8+6i\right)x=5+yi-\left(-3+4i\right)$$

Méadaigh -1 agus -3+4i chun 3-4i a fháil.

$$\left(8+6i\right)x=5+yi+\left(3-4i\right)$$

Déan suimiú in 5+\left(3-4i\right).

$$\left(8+6i\right)x=yi+8-4i$$

Tá an chothromóid i bhfoirm chaighdeánach.

$$\left(8+6i\right)x=iy+\left(8-4i\right)$$

Roinn an dá thaobh faoi 8+6i.

$$\frac{\left(8+6i\right)x}{8+6i}=\frac{iy+\left(8-4i\right)}{8+6i}$$

Má roinntear é faoi 8+6i cuirtear an iolrúchán faoi 8+6i ar ceal.

$$x=\frac{iy+\left(8-4i\right)}{8+6i}$$

Roinn iy+\left(8-4i\right) faoi 8+6i.

$$x=\left(\frac{3}{50}+\frac{2}{25}i\right)y+\left(\frac{2}{5}-\frac{4}{5}i\right)$$

Réitigh do y.

$$y=\left(6-8i\right)x+\left(4+8i\right)$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Úsáid an t-airí dáileach chun 2x+i a mhéadú faoi 4+3i.

$$\left(8+6i\right)x+\left(-3+4i\right)=5+yi$$

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

$$5+yi=\left(8+6i\right)x+\left(-3+4i\right)$$

Bain 5 ón dá thaobh.

$$yi=\left(8+6i\right)x+\left(-3+4i\right)-5$$

Déan suimiú in -3+4i-5.

$$yi=\left(8+6i\right)x-8+4i$$

Tá an chothromóid i bhfoirm chaighdeánach.

$$iy=\left(8+6i\right)x+\left(-8+4i\right)$$

Roinn an dá thaobh faoi i.

$$\frac{iy}{i}=\frac{\left(8+6i\right)x+\left(-8+4i\right)}{i}$$

Má roinntear é faoi i cuirtear an iolrúchán faoi i ar ceal.

$$y=\frac{\left(8+6i\right)x+\left(-8+4i\right)}{i}$$

Roinn \left(8+6i\right)x+\left(-8+4i\right) faoi i.

$$y=\left(6-8i\right)x+\left(4+8i\right)$$