Whakaoti i te {l}{ty+2=x}{x^2+4y^2=4}

$$\left\{ \begin{array} { l } { t y + 2 = x } \\ { x ^ { 2 } + 4 y ^ { 2 } = 4 } \end{array} \right.$$

Whakaoti mō x, y

$$x=2\text{, }y=0$$

$$x=\frac{2\left(4-t^{2}\right)}{t^{2}+4}\text{, }y=-\frac{4t}{t^{2}+4}$$

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

Whakaarohia te whārite tuatahi. Tangohia te x mai i ngā taha e rua.

$$ty+2-x=0$$

Tangohia te 2 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$ty-x=-2$$

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.

$$ty-x=-2,x^{2}+4y^{2}=4$$

Whakaotia te ty-x=-2 mō y mā te wehe i te y i te taha mauī o te tohu ōrite.

$$ty-x=-2$$

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

$$ty=x-2$$

Whakawehea ngā taha e rua ki te t.

$$y=\frac{1}{t}x-\frac{2}{t}$$

Whakakapia te \frac{1}{t}x-\frac{2}{t} mō te y ki tērā atu whārite, x^{2}+4y^{2}=4.

$$x^{2}+4\left(\frac{1}{t}x-\frac{2}{t}\right)^{2}=4$$

Pūrua \frac{1}{t}x-\frac{2}{t}.

$$x^{2}+4\left(\left(\frac{1}{t}\right)^{2}x^{2}+2\left(-\frac{2}{t}\right)\times \frac{1}{t}x+\left(-\frac{2}{t}\right)^{2}\right)=4$$

Whakareatia 4 ki te \left(\frac{1}{t}\right)^{2}x^{2}+2\left(-\frac{2}{t}\right)\times \frac{1}{t}x+\left(-\frac{2}{t}\right)^{2}.

$$x^{2}+4\times \left(\frac{1}{t}\right)^{2}x^{2}+8\left(-\frac{2}{t}\right)\times \frac{1}{t}x+4\left(-\frac{2}{t}\right)^{2}=4$$

Tāpiri x^{2} ki te 4\times \left(\frac{1}{t}\right)^{2}x^{2}.

$$\left(1+4\times \left(\frac{1}{t}\right)^{2}\right)x^{2}+8\left(-\frac{2}{t}\right)\times \frac{1}{t}x+4\left(-\frac{2}{t}\right)^{2}=4$$

Me tango 4 mai i ngā taha e rua o te whārite.

$$\left(1+4\times \left(\frac{1}{t}\right)^{2}\right)x^{2}+8\left(-\frac{2}{t}\right)\times \frac{1}{t}x+4\left(-\frac{2}{t}\right)^{2}-4=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1+4\times \left(\frac{1}{t}\right)^{2} mō a, 4\times 2\times \frac{1}{t}\left(-\frac{2}{t}\right) mō b, me \frac{16}{t^{2}}-4 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8\left(-\frac{2}{t}\right)\times \frac{1}{t}±\sqrt{\left(8\left(-\frac{2}{t}\right)\times \frac{1}{t}\right)^{2}-4\left(1+4\times \left(\frac{1}{t}\right)^{2}\right)\left(-4+\frac{16}{t^{2}}\right)}}{2\left(1+4\times \left(\frac{1}{t}\right)^{2}\right)}$$

Pūrua 4\times 2\times \frac{1}{t}\left(-\frac{2}{t}\right).

$$x=\frac{-8\left(-\frac{2}{t}\right)\times \frac{1}{t}±\sqrt{\frac{256}{t^{4}}-4\left(1+4\times \left(\frac{1}{t}\right)^{2}\right)\left(-4+\frac{16}{t^{2}}\right)}}{2\left(1+4\times \left(\frac{1}{t}\right)^{2}\right)}$$

Whakareatia -4 ki te 1+4\times \left(\frac{1}{t}\right)^{2}.

$$x=\frac{-8\left(-\frac{2}{t}\right)\times \frac{1}{t}±\sqrt{\frac{256}{t^{4}}+\left(-4-\frac{16}{t^{2}}\right)\left(-4+\frac{16}{t^{2}}\right)}}{2\left(1+4\times \left(\frac{1}{t}\right)^{2}\right)}$$

Whakareatia -4-\frac{16}{t^{2}} ki te \frac{16}{t^{2}}-4.

$$x=\frac{-8\left(-\frac{2}{t}\right)\times \frac{1}{t}±\sqrt{\frac{256}{t^{4}}+16-\frac{256}{t^{4}}}}{2\left(1+4\times \left(\frac{1}{t}\right)^{2}\right)}$$

Tāpiri \frac{256}{t^{4}} ki te -\frac{256}{t^{4}}+16.

$$x=\frac{-8\left(-\frac{2}{t}\right)\times \frac{1}{t}±\sqrt{16}}{2\left(1+4\times \left(\frac{1}{t}\right)^{2}\right)}$$

Tuhia te pūtakerua o te 16.

$$x=\frac{-8\left(-\frac{2}{t}\right)\times \frac{1}{t}±4}{2\left(1+4\times \left(\frac{1}{t}\right)^{2}\right)}$$

Whakareatia 2 ki te 1+4\times \left(\frac{1}{t}\right)^{2}.

$$x=\frac{\frac{16}{t^{2}}±4}{2+\frac{8}{t^{2}}}$$

Nā, me whakaoti te whārite x=\frac{\frac{16}{t^{2}}±4}{2+\frac{8}{t^{2}}} ina he tāpiri te ±. Tāpiri \frac{16}{t^{2}} ki te 4.

$$x=\frac{4+\frac{16}{t^{2}}}{2+\frac{8}{t^{2}}}$$

Whakawehe 4+\frac{16}{t^{2}} ki te 2+\frac{8}{t^{2}}.

$$x=2$$

Nā, me whakaoti te whārite x=\frac{\frac{16}{t^{2}}±4}{2+\frac{8}{t^{2}}} ina he tango te ±. Tango 4 mai i \frac{16}{t^{2}}.

$$x=\frac{-4+\frac{16}{t^{2}}}{2+\frac{8}{t^{2}}}$$

Whakawehe \frac{16}{t^{2}}-4 ki te 2+\frac{8}{t^{2}}.

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

E rua ngā otinga mō x: 2 me -\frac{2\left(t^{2}-4\right)}{4+t^{2}}. Me whakakapi 2 mō x ki te whārite y=\frac{1}{t}x-\frac{2}{t} hei kimi i te otinga hāngai mō y e pai ai ki ngā whārite e rua.

$$y=\frac{1}{t}\times 2-\frac{2}{t}$$

Whakareatia \frac{1}{t} ki te 2.

$$y=2\times \frac{1}{t}-\frac{2}{t}$$

Me whakakapi te -\frac{2\left(t^{2}-4\right)}{4+t^{2}} ināianei mō te x ki te whārite y=\frac{1}{t}x-\frac{2}{t} ka whakaoti hei kimi i te otinga hāngai mō y e pai ai ki ngā whārite e rua.

$$y=\frac{1}{t}\left(-\frac{2\left(t^{2}-4\right)}{t^{2}+4}\right)-\frac{2}{t}$$

Whakareatia \frac{1}{t} ki te -\frac{2\left(t^{2}-4\right)}{4+t^{2}}.

$$y=\left(-\frac{2\left(t^{2}-4\right)}{t^{2}+4}\right)\times \frac{1}{t}-\frac{2}{t}$$

Kua oti te pūnaha te whakatau.

$$y=2\times \frac{1}{t}-\frac{2}{t},x=2\text{ or }y=\left(-\frac{2\left(t^{2}-4\right)}{t^{2}+4}\right)\times \frac{1}{t}-\frac{2}{t},x=-\frac{2\left(t^{2}-4\right)}{t^{2}+4}$$

Whakaoti mō x, y (complex solution)

$$\left\{\begin{matrix}\\x=2\text{, }y=0\text{, }&\text{unconditionally}\\x=\frac{2\left(4-t^{2}\right)}{t^{2}+4}\text{, }y=-\frac{4t}{t^{2}+4}\text{, }&t\neq -2i\text{ and }t\neq 2i\end{matrix}\right.$$