Whakaoti i te (x+1)y^2=12

$$( x + 1 ) y ^ { 2 } = 12$$

Whakaoti mō x

$$x=-1+\frac{12}{y^{2}}$$

$$y\neq 0$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te āhuatanga tohatoha hei whakarea te x+1 ki te y^{2}.

$$xy^{2}+y^{2}=12$$

Tangohia te y^{2} mai i ngā taha e rua.

$$xy^{2}=12-y^{2}$$

He hanga arowhānui tō te whārite.

$$y^{2}x=12-y^{2}$$

Whakawehea ngā taha e rua ki te y^{2}.

$$\frac{y^{2}x}{y^{2}}=\frac{12-y^{2}}{y^{2}}$$

Mā te whakawehe ki te y^{2} ka wetekia te whakareanga ki te y^{2}.

$$x=\frac{12-y^{2}}{y^{2}}$$

Whakawehe 12-y^{2} ki te y^{2}.

$$x=-1+\frac{12}{y^{2}}$$

Whakaoti mō y (complex solution)

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

$$y=2\sqrt{3}\left(x+1\right)^{-\frac{1}{2}}\text{, }x\neq -1$$

Whakaoti mō y

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

$$y=-2\sqrt{\frac{3}{x+1}}\text{, }x>-1$$