Whakaoti i te y=2/3esqrt{2x-2}-1/3e

$$y = \frac { 2 } { 3 } e \sqrt { 2 x - 2 } - \frac { 1 } { 3 } e$$

Whakaoti mō x

$$x=\frac{9\times \left(\frac{y}{e}\right)^{2}}{8}+\frac{3y}{4e}+\frac{9}{8}$$

$$y\geq -\frac{e}{3}$$

Ngā Upane mō te Whakaoti Whārite Pūtake

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$\frac{2}{3}e\sqrt{2x-2}-\frac{1}{3}e=y$$

Me tāpiri te \frac{1}{3}e ki ngā taha e rua.

$$\frac{2}{3}e\sqrt{2x-2}=y+\frac{1}{3}e$$

Whakawehea ngā taha e rua ki te \frac{2}{3}e.

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

Mā te whakawehe ki te \frac{2}{3}e ka wetekia te whakareanga ki te \frac{2}{3}e.

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

Whakawehe y+\frac{e}{3} ki te \frac{2}{3}e.

$$\sqrt{2x-2}=\frac{3y}{2e}+\frac{1}{2}$$

Pūruatia ngā taha e rua o te whārite.

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

Me tāpiri 2 ki ngā taha e rua o te whārite.

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

Mā te tango i te -2 i a ia ake anō ka toe ko te 0.

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

Tango -2 mai i \frac{\left(e+3y\right)^{2}}{4e^{2}}.

$$2x=\frac{\frac{9y^{2}}{4}+\frac{3ey}{2}}{e^{2}}+\frac{9}{4}$$

Whakawehea ngā taha e rua ki te 2.

$$\frac{2x}{2}=\frac{\frac{\frac{9y^{2}}{4}+\frac{3ey}{2}}{e^{2}}+\frac{9}{4}}{2}$$

Mā te whakawehe ki te 2 ka wetekia te whakareanga ki te 2.

$$x=\frac{\frac{\frac{9y^{2}}{4}+\frac{3ey}{2}}{e^{2}}+\frac{9}{4}}{2}$$

Whakawehe \frac{9}{4}+\frac{\frac{9y^{2}}{4}+\frac{3ye}{2}}{e^{2}} ki te 2.

$$x=\frac{\frac{9y^{2}}{8}+\frac{3ey}{4}}{e^{2}}+\frac{9}{8}$$

Whakaoti mō x (complex solution)

$$x=\frac{9\times \left(\frac{y}{e}\right)^{2}}{8}+\frac{3y}{4e}+\frac{9}{8}$$

$$arg(3y+e)<\pi \text{ or }y=-\frac{e}{3}$$

Whakaoti mō y (complex solution)

$$y=\frac{e\left(2\sqrt{2\left(x-1\right)}-1\right)}{3}$$

Whakaoti mō y

$$y=\frac{e\left(2\sqrt{2\left(x-1\right)}-1\right)}{3}$$

$$x\geq 1$$