Whakaoti i te 133/v-1/40=133-1/-20

$$\frac { 133 } { v } - \frac { 1 } { 40 } = \frac { 133 - 1 } { - 20 }$$

Whakaoti mō v

$$v = -\frac{5320}{263} = -20\frac{60}{263} \approx -20.228136882$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Tē taea kia ōrite te tāupe v ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te 40v, arā, te tauraro pātahi he tino iti rawa te kitea o v,40,-20.

$$40\times 133+40v\left(-\frac{1}{40}\right)=-2v\left(133-1\right)$$

Whakareatia te 40 ki te 133, ka 5320.

$$5320+40v\left(-\frac{1}{40}\right)=-2v\left(133-1\right)$$

Me whakakore te 40 me te 40.

$$5320-v=-2v\left(133-1\right)$$

Tangohia te 1 i te 133, ka 132.

$$5320-v=-2v\times 132$$

Whakareatia te -2 ki te 132, ka -264.

$$5320-v=-264v$$

Me tāpiri te 264v ki ngā taha e rua.

$$5320-v+264v=0$$

Pahekotia te -v me 264v, ka 263v.

$$5320+263v=0$$

Tangohia te 5320 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$263v=-5320$$

Whakawehea ngā taha e rua ki te 263.

$$v=\frac{-5320}{263}$$

Ka taea te hautanga \frac{-5320}{263} te tuhi anō ko -\frac{5320}{263} mā te tango i te tohu tōraro.

$$v=-\frac{5320}{263}$$