Whakaoti i te F+2=H-10k(E+2)

$$F + 2 = H - 10 k ( E + 2 )$$

Whakaoti mō E (complex solution)

$$\left\{\begin{matrix}E=\frac{-F+H-20k-2}{10k}\text{, }&k\neq 0\\E\in \mathrm{C}\text{, }&F=H-2\text{ and }k=0\end{matrix}\right.$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$H-10k\left(E+2\right)=F+2$$

Whakamahia te āhuatanga tohatoha hei whakarea te -10k ki te E+2.

$$H-10kE-20k=F+2$$

Tangohia te H mai i ngā taha e rua.

$$-10kE-20k=F+2-H$$

Me tāpiri te 20k ki ngā taha e rua.

$$-10kE=F+2-H+20k$$

He hanga arowhānui tō te whārite.

$$\left(-10k\right)E=F-H+20k+2$$

Whakawehea ngā taha e rua ki te -10k.

$$\frac{\left(-10k\right)E}{-10k}=\frac{F-H+20k+2}{-10k}$$

Mā te whakawehe ki te -10k ka wetekia te whakareanga ki te -10k.

$$E=\frac{F-H+20k+2}{-10k}$$

Whakawehe F-H+2+20k ki te -10k.

$$E=-\frac{F-H+20k+2}{10k}$$

Whakaoti mō E

$$\left\{\begin{matrix}E=\frac{-F+H-20k-2}{10k}\text{, }&k\neq 0\\E\in \mathrm{R}\text{, }&F=H-2\text{ and }k=0\end{matrix}\right.$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$H-10k\left(E+2\right)=F+2$$

Whakamahia te āhuatanga tohatoha hei whakarea te -10k ki te E+2.

$$H-10kE-20k=F+2$$

Tangohia te H mai i ngā taha e rua.

$$-10kE-20k=F+2-H$$

Me tāpiri te 20k ki ngā taha e rua.

$$-10kE=F+2-H+20k$$

He hanga arowhānui tō te whārite.

$$\left(-10k\right)E=F-H+20k+2$$

Whakawehea ngā taha e rua ki te -10k.

$$\frac{\left(-10k\right)E}{-10k}=\frac{F-H+20k+2}{-10k}$$

Mā te whakawehe ki te -10k ka wetekia te whakareanga ki te -10k.

$$E=\frac{F-H+20k+2}{-10k}$$

Whakawehe F-H+2+20k ki te -10k.

$$E=-\frac{F-H+20k+2}{10k}$$

Whakaoti mō F

$$F=-10Ek+H-20k-2$$

Ngā Upane Otinga

Tangohia te 2 mai i ngā taha e rua.

$$F=H-10k\left(E+2\right)-2$$

Whakamahia te āhuatanga tohatoha hei whakarea te -10k ki te E+2.

$$F=H-10kE-20k-2$$