Whakaoti i te x/c-3=cx+6

$$\frac { x } { c - 3 } = c x + 6$$

Whakaoti mō x

$$x=-\frac{6\left(3-c\right)}{1+3c-c^{2}}$$

$$c\neq \frac{\sqrt{13}+3}{2}\text{ and }c\neq \frac{3-\sqrt{13}}{2}\text{ and }c\neq 3$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakareatia ngā taha e rua o te whārite ki te c-3.

$$x=cx\left(c-3\right)+\left(c-3\right)\times 6$$

Whakamahia te āhuatanga tohatoha hei whakarea te cx ki te c-3.

$$x=xc^{2}-3cx+\left(c-3\right)\times 6$$

Whakamahia te āhuatanga tohatoha hei whakarea te c-3 ki te 6.

$$x=xc^{2}-3cx+6c-18$$

Tangohia te xc^{2} mai i ngā taha e rua.

$$x-xc^{2}=-3cx+6c-18$$

Me tāpiri te 3cx ki ngā taha e rua.

$$x-xc^{2}+3cx=6c-18$$

Whakaraupapatia anō ngā kīanga tau.

$$-xc^{2}+3cx+x=6c-18$$

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

$$\left(-c^{2}+3c+1\right)x=6c-18$$

He hanga arowhānui tō te whārite.

$$\left(1+3c-c^{2}\right)x=6c-18$$

Whakawehea ngā taha e rua ki te -c^{2}+3c+1.

$$\frac{\left(1+3c-c^{2}\right)x}{1+3c-c^{2}}=\frac{6c-18}{1+3c-c^{2}}$$

Mā te whakawehe ki te -c^{2}+3c+1 ka wetekia te whakareanga ki te -c^{2}+3c+1.

$$x=\frac{6c-18}{1+3c-c^{2}}$$

Whakawehe -18+6c ki te -c^{2}+3c+1.

$$x=\frac{6\left(c-3\right)}{1+3c-c^{2}}$$

Whakaoti mō c

$$c=-\frac{\sqrt{13x^{2}+36x+36}-3x+6}{2x}$$

$$c=-\frac{-\sqrt{13x^{2}+36x+36}-3x+6}{2x}\text{, }x\neq 0$$