Whakaoti i te a(1-q^3)/1-q=12

$$\frac { a ( 1 - q ^ { 3 } ) } { 1 - q } = 12$$

Whakaoti mō a

$$a=\frac{12}{q^{2}+q+1}$$

$$q\neq 1$$

Ngā Upane mō te Whakaoti Whārite Rārangi

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

$$a\left(1-q^{3}\right)=12\left(-q+1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te a ki te 1-q^{3}.

$$a-aq^{3}=12\left(-q+1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 12 ki te -q+1.

$$a-aq^{3}=-12q+12$$

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

$$\left(1-q^{3}\right)a=-12q+12$$

He hanga arowhānui tō te whārite.

$$\left(1-q^{3}\right)a=12-12q$$

Whakawehea ngā taha e rua ki te 1-q^{3}.

$$\frac{\left(1-q^{3}\right)a}{1-q^{3}}=\frac{12-12q}{1-q^{3}}$$

Mā te whakawehe ki te 1-q^{3} ka wetekia te whakareanga ki te 1-q^{3}.

$$a=\frac{12-12q}{1-q^{3}}$$

Whakawehe -12q+12 ki te 1-q^{3}.

$$a=\frac{12}{q^{2}+q+1}$$

Whakaoti mō q

$$\left\{\begin{matrix}q=\frac{-\sqrt{-3+\frac{48}{a}}-1}{2}\text{, }&a>0\text{ and }a\leq 16\\q=\frac{\sqrt{-3+\frac{48}{a}}-1}{2}\text{, }&a\neq 4\text{ and }a\leq 16\text{ and }a>0\end{matrix}\right.$$