Whakaoti i te 1/x+1/t=1

$$\frac{ 1 }{ x } + \frac{ 1 }{ t } = 1$$

Whakaoti mō t

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

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

Ngā Upane mō te Whakaoti Whārite Rārangi

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

$$t+x=tx$$

Tangohia te tx mai i ngā taha e rua.

$$t+x-tx=0$$

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

$$t-tx=-x$$

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

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

Whakawehea ngā taha e rua ki te 1-x.

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

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

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

Tē taea kia ōrite te tāupe t ki 0.

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

Whakaoti mō x

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

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

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 tx, arā, te tauraro pātahi he tino iti rawa te kitea o x,t.

$$t+x=tx$$

Tangohia te tx mai i ngā taha e rua.

$$t+x-tx=0$$

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

$$x-tx=-t$$

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

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

Whakawehea ngā taha e rua ki te 1-t.

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

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

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

Tē taea kia ōrite te tāupe x ki 0.

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