Whakaoti i te sqrt{3/4x-6}=y

$$\sqrt { \frac { 3 } { 4 } x - 6 } = y$$

Whakaoti mō x

$$x=\frac{4\left(y^{2}+6\right)}{3}$$

$$y\geq 0$$

Ngā Upane mō te Whakaoti Whārite Pūtake

Pūruatia ngā taha e rua o te whārite.

$$\frac{3}{4}x-6=y^{2}$$

Me tāpiri 6 ki ngā taha e rua o te whārite.

$$\frac{3}{4}x-6-\left(-6\right)=y^{2}-\left(-6\right)$$

Mā te tango i te -6 i a ia ake anō ka toe ko te 0.

$$\frac{3}{4}x=y^{2}-\left(-6\right)$$

Tango -6 mai i y^{2}.

$$\frac{3}{4}x=y^{2}+6$$

Whakawehea ngā taha e rua o te whārite ki te \frac{3}{4}, he ōrite ki te whakarea i ngā taha e rua ki te tau huripoki o te hautanga.

$$\frac{\frac{3}{4}x}{\frac{3}{4}}=\frac{y^{2}+6}{\frac{3}{4}}$$

Mā te whakawehe ki te \frac{3}{4} ka wetekia te whakareanga ki te \frac{3}{4}.

$$x=\frac{y^{2}+6}{\frac{3}{4}}$$

Whakawehe y^{2}+6 ki te \frac{3}{4} mā te whakarea y^{2}+6 ki te tau huripoki o \frac{3}{4}.

$$x=\frac{4y^{2}}{3}+8$$

Whakaoti mō x (complex solution)

$$x=\frac{4\left(y^{2}+6\right)}{3}$$

$$arg(y)<\pi \text{ or }y=0$$

Whakaoti mō y (complex solution)

$$y=\frac{\sqrt{3\left(x-8\right)}}{2}$$

Whakaoti mō y

$$y=\frac{\sqrt{3\left(x-8\right)}}{2}$$

$$x\geq 8$$