Whakaoti i te 1/x+1/a=1.xneq0text{i}aneq0

$$\frac { 1 } { x } + \frac { 1 } { a } = 1 . x \neq 0 \text { i } a \neq 0$$

Whakaoti mō a

$$a=-\frac{x}{1-x}$$

$$x\neq 0\text{ and }x\neq 1$$

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 ax, arā, te tauraro pātahi he tino iti rawa te kitea o x,a.

$$a+x=ax$$

Tangohia te ax mai i ngā taha e rua.

$$a+x-ax=0$$

Tangohia te x mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$a-ax=-x$$

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

$$\left(1-x\right)a=-x$$

Whakawehea ngā taha e rua ki te 1-x.

$$\frac{\left(1-x\right)a}{1-x}=-\frac{x}{1-x}$$

Mā te whakawehe ki te 1-x ka wetekia te whakareanga ki te 1-x.

$$a=-\frac{x}{1-x}$$

Tē taea kia ōrite te tāupe a ki 0.

$$a=-\frac{x}{1-x}\text{, }a\neq 0$$

Whakaoti mō x

$$x=-\frac{a}{1-a}$$

$$a\neq 0\text{ and }a\neq 1$$

Ngā Upane mō te Whakaoti Whārite Rārangi

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

$$a+x=ax$$

Tangohia te ax mai i ngā taha e rua.

$$a+x-ax=0$$

Tangohia te a mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$x-ax=-a$$

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

$$\left(1-a\right)x=-a$$

Whakawehea ngā taha e rua ki te 1-a.

$$\frac{\left(1-a\right)x}{1-a}=-\frac{a}{1-a}$$

Mā te whakawehe ki te 1-a ka wetekia te whakareanga ki te 1-a.

$$x=-\frac{a}{1-a}$$

Tē taea kia ōrite te tāupe x ki 0.

$$x=-\frac{a}{1-a}\text{, }x\neq 0$$