Whakaoti i te 2/50^2+1/30b=6ab+5a^2/15a^3b=

$$\frac { 2 } { 50 ^ { 2 } } + \frac { 1 } { 30 b } = \frac { 6 a b + 5 a ^ { 2 } } { 15 a ^ { 3 } b } =$$

Whakaoti mō b (complex solution)

$$b=-\frac{125a\left(a-10\right)}{3\left(a^{2}-500\right)}$$

$$a\neq 10\text{ and }a\neq 0\text{ and }a\neq -10\sqrt{5}\text{ and }a\neq 10\sqrt{5}$$

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 30ba^{3}, arā, te tauraro pātahi he tino iti rawa te kitea o 30b,15a^{3}b.

$$\frac{3}{250}ba^{3}\times 2+a^{3}=2\left(6ab+5a^{2}\right)$$

Whakareatia te \frac{3}{250} ki te 2, ka \frac{3}{125}.

$$\frac{3}{125}ba^{3}+a^{3}=2\left(6ab+5a^{2}\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 2 ki te 6ab+5a^{2}.

$$\frac{3}{125}ba^{3}+a^{3}=12ab+10a^{2}$$

Tangohia te 12ab mai i ngā taha e rua.

$$\frac{3}{125}ba^{3}+a^{3}-12ab=10a^{2}$$

Tangohia te a^{3} mai i ngā taha e rua.

$$\frac{3}{125}ba^{3}-12ab=10a^{2}-a^{3}$$

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

$$\left(\frac{3}{125}a^{3}-12a\right)b=10a^{2}-a^{3}$$

He hanga arowhānui tō te whārite.

$$\left(\frac{3a^{3}}{125}-12a\right)b=10a^{2}-a^{3}$$

Whakawehea ngā taha e rua ki te \frac{3}{125}a^{3}-12a.

$$\frac{\left(\frac{3a^{3}}{125}-12a\right)b}{\frac{3a^{3}}{125}-12a}=\frac{\left(10-a\right)a^{2}}{\frac{3a^{3}}{125}-12a}$$

Mā te whakawehe ki te \frac{3}{125}a^{3}-12a ka wetekia te whakareanga ki te \frac{3}{125}a^{3}-12a.

$$b=\frac{\left(10-a\right)a^{2}}{\frac{3a^{3}}{125}-12a}$$

Whakawehe \left(10-a\right)a^{2} ki te \frac{3}{125}a^{3}-12a.

$$b=\frac{125a\left(10-a\right)}{3\left(a^{2}-500\right)}$$

Tē taea kia ōrite te tāupe b ki 0.

$$b=\frac{125a\left(10-a\right)}{3\left(a^{2}-500\right)}\text{, }b\neq 0$$

Whakaoti mō b

$$b=-\frac{125a\left(a-10\right)}{3\left(a^{2}-500\right)}$$

$$a\neq 10\text{ and }a\neq 0\text{ and }|a|\neq 10\sqrt{5}$$

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 30ba^{3}, arā, te tauraro pātahi he tino iti rawa te kitea o 30b,15a^{3}b.

$$\frac{3}{250}ba^{3}\times 2+a^{3}=2\left(6ab+5a^{2}\right)$$

Whakareatia te \frac{3}{250} ki te 2, ka \frac{3}{125}.

$$\frac{3}{125}ba^{3}+a^{3}=2\left(6ab+5a^{2}\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 2 ki te 6ab+5a^{2}.

$$\frac{3}{125}ba^{3}+a^{3}=12ab+10a^{2}$$

Tangohia te 12ab mai i ngā taha e rua.

$$\frac{3}{125}ba^{3}+a^{3}-12ab=10a^{2}$$

Tangohia te a^{3} mai i ngā taha e rua.

$$\frac{3}{125}ba^{3}-12ab=10a^{2}-a^{3}$$

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

$$\left(\frac{3}{125}a^{3}-12a\right)b=10a^{2}-a^{3}$$

He hanga arowhānui tō te whārite.

$$\left(\frac{3a^{3}}{125}-12a\right)b=10a^{2}-a^{3}$$

Whakawehea ngā taha e rua ki te \frac{3}{125}a^{3}-12a.

$$\frac{\left(\frac{3a^{3}}{125}-12a\right)b}{\frac{3a^{3}}{125}-12a}=\frac{\left(10-a\right)a^{2}}{\frac{3a^{3}}{125}-12a}$$

Mā te whakawehe ki te \frac{3}{125}a^{3}-12a ka wetekia te whakareanga ki te \frac{3}{125}a^{3}-12a.

$$b=\frac{\left(10-a\right)a^{2}}{\frac{3a^{3}}{125}-12a}$$

Whakawehe \left(10-a\right)a^{2} ki te \frac{3}{125}a^{3}-12a.

$$b=\frac{125a\left(10-a\right)}{3\left(a^{2}-500\right)}$$

Tē taea kia ōrite te tāupe b ki 0.

$$b=\frac{125a\left(10-a\right)}{3\left(a^{2}-500\right)}\text{, }b\neq 0$$

Whakaoti mō a (complex solution)

$$\left\{\begin{matrix}a=\frac{5\left(\sqrt{5\left(36b^{2}+1500b+3125\right)}+125\right)}{3b+125}\text{; }a=\frac{5\left(-\sqrt{5\left(36b^{2}+1500b+3125\right)}+125\right)}{3b+125}\text{, }&b\neq 0\text{ and }b\neq -\frac{125}{3}\\a=50\text{, }&b=-\frac{125}{3}\end{matrix}\right.$$

Whakaoti mō a

$$\left\{\begin{matrix}a=\frac{5\left(\sqrt{5\left(36b^{2}+1500b+3125\right)}+125\right)}{3b+125}\text{; }a=\frac{5\left(-\sqrt{5\left(36b^{2}+1500b+3125\right)}+125\right)}{3b+125}\text{, }&\left(b\neq 0\text{ and }b\geq \frac{25\sqrt{5}}{3}-\frac{125}{6}\right)\text{ or }\left(b\neq -\frac{125}{3}\text{ and }b\leq -\frac{25\sqrt{5}}{3}-\frac{125}{6}\right)\\a=50\text{, }&b=-\frac{125}{3}\end{matrix}\right.$$