Whakaoti i te x/a-1=ax+4

$$\frac { x } { a - 1 } = a x + 4$$

Whakaoti mō x

$$x=-\frac{4\left(1-a\right)}{1+a-a^{2}}$$

$$a\neq \frac{\sqrt{5}+1}{2}\text{ and }a\neq \frac{1-\sqrt{5}}{2}\text{ and }a\neq 1$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakareatia ngā taha e rua o te whārite ki te a-1.

$$x=ax\left(a-1\right)+\left(a-1\right)\times 4$$

Whakamahia te āhuatanga tohatoha hei whakarea te ax ki te a-1.

$$x=xa^{2}-ax+\left(a-1\right)\times 4$$

Whakamahia te āhuatanga tohatoha hei whakarea te a-1 ki te 4.

$$x=xa^{2}-ax+4a-4$$

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

$$x-xa^{2}=-ax+4a-4$$

Me tāpiri te ax ki ngā taha e rua.

$$x-xa^{2}+ax=4a-4$$

Whakaraupapatia anō ngā kīanga tau.

$$ax-xa^{2}+x=4a-4$$

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

$$\left(a-a^{2}+1\right)x=4a-4$$

He hanga arowhānui tō te whārite.

$$\left(1+a-a^{2}\right)x=4a-4$$

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

$$\frac{\left(1+a-a^{2}\right)x}{1+a-a^{2}}=\frac{4a-4}{1+a-a^{2}}$$

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

$$x=\frac{4a-4}{1+a-a^{2}}$$

Whakawehe -4+4a ki te 1-a^{2}+a.

$$x=\frac{4\left(a-1\right)}{1+a-a^{2}}$$

Whakaoti mō a

$$a=-\frac{\sqrt{5x^{2}+8x+16}-x+4}{2x}$$

$$a=-\frac{-\sqrt{5x^{2}+8x+16}-x+4}{2x}\text{, }x\neq 0$$