Whakaoti i te y=2-sqrt{5-x}

$$y = 2 - \sqrt { 5 - x }$$

Whakaoti mō x

$$x=-\left(2-y\right)^{2}+5$$

$$2-y\geq 0$$

Ngā Upane mō te Whakaoti Whārite Pūtake

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

$$2-\sqrt{5-x}=y$$

Tangohia te 2 mai i ngā taha e rua.

$$-\sqrt{5-x}=y-2$$

Whakawehea ngā taha e rua ki te -1.

$$\frac{-\sqrt{-x+5}}{-1}=\frac{y-2}{-1}$$

Mā te whakawehe ki te -1 ka wetekia te whakareanga ki te -1.

$$\sqrt{-x+5}=\frac{y-2}{-1}$$

Whakawehe y-2 ki te -1.

$$\sqrt{-x+5}=2-y$$

Pūruatia ngā taha e rua o te whārite.

$$-x+5=\left(2-y\right)^{2}$$

Me tango 5 mai i ngā taha e rua o te whārite.

$$-x+5-5=\left(2-y\right)^{2}-5$$

Mā te tango i te 5 i a ia ake anō ka toe ko te 0.

$$-x=\left(2-y\right)^{2}-5$$

Whakawehea ngā taha e rua ki te -1.

$$\frac{-x}{-1}=\frac{\left(2-y\right)^{2}-5}{-1}$$

Mā te whakawehe ki te -1 ka wetekia te whakareanga ki te -1.

$$x=\frac{\left(2-y\right)^{2}-5}{-1}$$

Whakawehe \left(-y+2\right)^{2}-5 ki te -1.

$$x=-\left(2-y\right)^{2}+5$$

Whakaoti mō x (complex solution)

$$x=-\left(2-y\right)^{2}+5$$

$$y=2\text{ or }arg(2-y)<\pi$$

Whakaoti mō y (complex solution)

$$y=-\sqrt{5-x}+2$$

Whakaoti mō y

$$y=-\sqrt{5-x}+2$$

$$x\leq 5$$