Whakaoti i te sqrt{2x}+y-1=5

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

Whakaoti mō x

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

$$6-y\geq 0$$

Ngā Upane mō te Whakaoti Whārite Pūtake

Me tango y-1 mai i ngā taha e rua o te whārite.

$$\sqrt{2x}+y-1-\left(y-1\right)=5-\left(y-1\right)$$

Mā te tango i te y-1 i a ia ake anō ka toe ko te 0.

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

Tango y-1 mai i 5.

$$\sqrt{2x}=6-y$$

Pūruatia ngā taha e rua o te whārite.

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

Whakawehea ngā taha e rua ki te 2.

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

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

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

Whakaoti mō x (complex solution)

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

$$y=6\text{ or }arg(6-y)<\pi$$

Whakaoti mō y (complex solution)

$$y=-\sqrt{2x}+6$$

Whakaoti mō y

$$y=-\sqrt{2x}+6$$

$$x\geq 0$$