Whakaoti i te 1/x=1/a+1/b

$$\frac { 1 } { x } = \frac { 1 } { a } + \frac { 1 } { b }$$

Whakaoti mō a

$$a=-\frac{bx}{x-b}$$

$$b\neq 0\text{ and }x\neq 0\text{ and }x\neq b$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Tē taea kia ōrite te tāupe a ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te abx, arā, te tauraro pātahi he tino iti rawa te kitea o x,a,b.

$$ab=bx+ax$$

Tangohia te ax mai i ngā taha e rua.

$$ab-ax=bx$$

Pahekotia ngā kīanga tau katoa e whai ana i te a.

$$\left(b-x\right)a=bx$$

Whakawehea ngā taha e rua ki te b-x.

$$\frac{\left(b-x\right)a}{b-x}=\frac{bx}{b-x}$$

Mā te whakawehe ki te b-x ka wetekia te whakareanga ki te b-x.

$$a=\frac{bx}{b-x}$$

Tē taea kia ōrite te tāupe a ki 0.

$$a=\frac{bx}{b-x}\text{, }a\neq 0$$

Whakaoti mō b

$$b=-\frac{ax}{x-a}$$

$$a\neq 0\text{ and }x\neq 0\text{ and }x\neq a$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Tē taea kia ōrite te tāupe b ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te abx, arā, te tauraro pātahi he tino iti rawa te kitea o x,a,b.

$$ab=bx+ax$$

Tangohia te bx mai i ngā taha e rua.

$$ab-bx=ax$$

Pahekotia ngā kīanga tau katoa e whai ana i te b.

$$\left(a-x\right)b=ax$$

Whakawehea ngā taha e rua ki te a-x.

$$\frac{\left(a-x\right)b}{a-x}=\frac{ax}{a-x}$$

Mā te whakawehe ki te a-x ka wetekia te whakareanga ki te a-x.

$$b=\frac{ax}{a-x}$$

Tē taea kia ōrite te tāupe b ki 0.

$$b=\frac{ax}{a-x}\text{, }b\neq 0$$