Whakaoti i te 3/AD+1/AB=1/8

$$\frac { 3 } { A D } + \frac { 1 } { A B } = \frac { 1 } { 8 }$$

Whakaoti mō A

$$A=\frac{8\left(3B+D\right)}{BD}$$

$$D\neq -3B\text{ and }D\neq 0\text{ and }B\neq 0$$

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 8ABD, arā, te tauraro pātahi he tino iti rawa te kitea o AD,AB,8.

$$8B\times 3+8D=ABD$$

Whakareatia te 8 ki te 3, ka 24.

$$24B+8D=ABD$$

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

$$ABD=24B+8D$$

He hanga arowhānui tō te whārite.

$$BDA=24B+8D$$

Whakawehea ngā taha e rua ki te BD.

$$\frac{BDA}{BD}=\frac{24B+8D}{BD}$$

Mā te whakawehe ki te BD ka wetekia te whakareanga ki te BD.

$$A=\frac{24B+8D}{BD}$$

Whakawehe 24B+8D ki te BD.

$$A=\frac{8}{B}+\frac{24}{D}$$

Tē taea kia ōrite te tāupe A ki 0.

$$A=\frac{8}{B}+\frac{24}{D}\text{, }A\neq 0$$

Whakaoti mō B

$$B=-\frac{8D}{24-AD}$$

$$D\neq 0\text{ and }A\neq 0\text{ and }A\neq \frac{24}{D}$$

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 8ABD, arā, te tauraro pātahi he tino iti rawa te kitea o AD,AB,8.

$$8B\times 3+8D=ABD$$

Whakareatia te 8 ki te 3, ka 24.

$$24B+8D=ABD$$

Tangohia te ABD mai i ngā taha e rua.

$$24B+8D-ABD=0$$

Tangohia te 8D mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$24B-ABD=-8D$$

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

$$\left(24-AD\right)B=-8D$$

Whakawehea ngā taha e rua ki te 24-AD.

$$\frac{\left(24-AD\right)B}{24-AD}=-\frac{8D}{24-AD}$$

Mā te whakawehe ki te 24-AD ka wetekia te whakareanga ki te 24-AD.

$$B=-\frac{8D}{24-AD}$$

Tē taea kia ōrite te tāupe B ki 0.

$$B=-\frac{8D}{24-AD}\text{, }B\neq 0$$