Whakaoti i te {l}{x^2+y^2=1}{x+y=1}

$$\left\{ \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 1 } \\ { x + y = 1 } \end{array} \right.$$

Whakaoti mō x, y

$$x=0\text{, }y=1$$

$$x=1\text{, }y=0$$

Ngā Upane e Whakamahi ana i te Whakakapinga me te Tikanga Tātai Pūrua

Hei whakaoti i ētahi whārite takirua mā te whakakapinga, me whakaoti tētahi whārite i te tuatahi mō tētahi o ngā taurangi. Ka whakakapi i te otinga mō taua taurangi ki tērā o ngā whārite.

$$x+y=1,y^{2}+x^{2}=1$$

Whakaotia te x+y=1 mō x mā te wehe i te x i te taha mauī o te tohu ōrite.

$$x+y=1$$

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

$$x=-y+1$$

Whakakapia te -y+1 mō te x ki tērā atu whārite, y^{2}+x^{2}=1.

$$y^{2}+\left(-y+1\right)^{2}=1$$

Pūrua -y+1.

$$y^{2}+y^{2}-2y+1=1$$

Tāpiri y^{2} ki te y^{2}.

$$2y^{2}-2y+1=1$$

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

$$2y^{2}-2y=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1+1\left(-1\right)^{2} mō a, 1\times 1\left(-1\right)\times 2 mō b, me 0 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}}}{2\times 2}$$

Tuhia te pūtakerua o te \left(-2\right)^{2}.

$$y=\frac{-\left(-2\right)±2}{2\times 2}$$

Ko te tauaro o 1\times 1\left(-1\right)\times 2 ko 2.

$$y=\frac{2±2}{2\times 2}$$

Whakareatia 2 ki te 1+1\left(-1\right)^{2}.

$$y=\frac{2±2}{4}$$

Nā, me whakaoti te whārite y=\frac{2±2}{4} ina he tāpiri te ±. Tāpiri 2 ki te 2.

$$y=\frac{4}{4}$$

Whakawehe 4 ki te 4.

$$y=1$$

Nā, me whakaoti te whārite y=\frac{2±2}{4} ina he tango te ±. Tango 2 mai i 2.

$$y=\frac{0}{4}$$

Whakawehe 0 ki te 4.

$$y=0$$

E rua ngā otinga mō y: 1 me 0. Me whakakapi 1 mō y ki te whārite x=-y+1 hei kimi i te otinga hāngai mō x e pai ai ki ngā whārite e rua.

$$x=-1+1$$

Tāpiri -1 ki te 1.

$$x=0$$

Me whakakapi te 0 ināianei mō te y ki te whārite x=-y+1 ka whakaoti hei kimi i te otinga hāngai mō x e pai ai ki ngā whārite e rua.

$$x=1$$

Kua oti te pūnaha te whakatau.

$$x=0,y=1\text{ or }x=1,y=0$$