Whakaoti i te {l}{ax-y=3}{a-4x+sqrt{2}=y}

$$\left. \begin{array} { l } { a x - y = 3 } \\ { a - 4 x + \sqrt { 2 } = y } \end{array} \right.$$

Whakaoti mō x, y

$$x=\frac{a+\sqrt{2}+3}{a+4}$$

$$y=\frac{a^{2}+\sqrt{2}a-12}{a+4}$$

$$a\neq -4$$

Ngā Upane e Whakamahi ana i te Whakakapinga

Whakaarohia te whārite tuarua. Tangohia te y mai i ngā taha e rua.

$$a-4x+\sqrt{2}-y=0$$

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

$$-4x+\sqrt{2}-y=-a$$

Tangohia te \sqrt{2} mai i ngā taha e rua.

$$-4x-y=-a-\sqrt{2}$$

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.

$$ax-y=3,-4x-y=-a-\sqrt{2}$$

Kōwhiria tētahi o ngā whārite ka whakaotia mō te x mā te wehe i te x i te taha mauī o te tohu ōrite.

$$ax-y=3$$

Me tāpiri y ki ngā taha e rua o te whārite.

$$ax=y+3$$

Whakawehea ngā taha e rua ki te a.

$$x=\frac{1}{a}\left(y+3\right)$$

Whakareatia \frac{1}{a} ki te y+3.

$$x=\frac{1}{a}y+\frac{3}{a}$$

Whakakapia te \frac{3+y}{a} mō te x ki tērā atu whārite, -4x-y=-a-\sqrt{2}.

$$-4\left(\frac{1}{a}y+\frac{3}{a}\right)-y=-a-\sqrt{2}$$

Whakareatia -4 ki te \frac{3+y}{a}.

$$\left(-\frac{4}{a}\right)y-\frac{12}{a}-y=-a-\sqrt{2}$$

Tāpiri -\frac{4y}{a} ki te -y.

$$\left(-1-\frac{4}{a}\right)y-\frac{12}{a}=-a-\sqrt{2}$$

Me tāpiri \frac{12}{a} ki ngā taha e rua o te whārite.

$$\left(-1-\frac{4}{a}\right)y=-a-\sqrt{2}+\frac{12}{a}$$

Whakawehea ngā taha e rua ki te -\frac{4}{a}-1.

$$y=-\frac{-a^{2}-\sqrt{2}a+12}{a+4}$$

Whakaurua te -\frac{12-\sqrt{2}a-a^{2}}{4+a} mō y ki x=\frac{1}{a}y+\frac{3}{a}. I te mea kotahi anake te taurangi kei te whārite i puta, ka taea e koe te whakaoti mō x hāngai tonu.

$$x=\frac{1}{a}\left(-\frac{-a^{2}-\sqrt{2}a+12}{a+4}\right)+\frac{3}{a}$$

Whakareatia \frac{1}{a} ki te -\frac{12-\sqrt{2}a-a^{2}}{4+a}.

$$x=-\frac{-a^{2}-\sqrt{2}a+12}{a\left(a+4\right)}+\frac{3}{a}$$

Tāpiri \frac{3}{a} ki te -\frac{12-\sqrt{2}a-a^{2}}{a\left(4+a\right)}.

$$x=\frac{a+\sqrt{2}+3}{a+4}$$

Kua oti te pūnaha te whakatau.

$$x=\frac{a+\sqrt{2}+3}{a+4},y=-\frac{-a^{2}-\sqrt{2}a+12}{a+4}$$

Ngā Upane e Whakamahi i te Tangohanga

Whakaarohia te whārite tuarua. Tangohia te y mai i ngā taha e rua.

$$a-4x+\sqrt{2}-y=0$$

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

$$-4x+\sqrt{2}-y=-a$$

Tangohia te \sqrt{2} mai i ngā taha e rua.

$$-4x-y=-a-\sqrt{2}$$

Hei whakaoti mā te tangohanga, ko ngā tau whakarea o tētahi o ngā taurangi me mātua ōrite i ngā whārite e rua kia whakakorehia ai te taurangi ina tangohia tētahi whārite mai i tētahi atu.

$$ax-y=3,-4x-y=-a-\sqrt{2}$$

Me tango -4x-y=-a-\sqrt{2} mai i ax-y=3 mā te tango i ngā kīanga tau ōrite i ia taha o te tohu ōrite.

$$ax+4x-y+y=3+a+\sqrt{2}$$

Tāpiri -y ki te y. Ka whakakore atu ngā kupu -y me y, ka toe he whārite me tētahi taurangi kotahi ka taea te whakaoti.

$$ax+4x=3+a+\sqrt{2}$$

Tāpiri ax ki te 4x.

$$\left(a+4\right)x=3+a+\sqrt{2}$$

Tāpiri 3 ki te a+\sqrt{2}.

$$\left(a+4\right)x=a+\sqrt{2}+3$$

Whakawehea ngā taha e rua ki te a+4.

$$x=\frac{a+\sqrt{2}+3}{a+4}$$

Whakaurua te \frac{3+a+\sqrt{2}}{a+4} mō x ki -4x-y=-a-\sqrt{2}. I te mea kotahi anake te taurangi kei te whārite i puta, ka taea e koe te whakaoti mō y hāngai tonu.

$$-4\times \frac{a+\sqrt{2}+3}{a+4}-y=-a-\sqrt{2}$$

Whakareatia -4 ki te \frac{3+a+\sqrt{2}}{a+4}.

$$-\frac{4\left(a+\sqrt{2}+3\right)}{a+4}-y=-a-\sqrt{2}$$

Me tāpiri \frac{4\left(3+a+\sqrt{2}\right)}{a+4} ki ngā taha e rua o te whārite.

$$-y=\frac{-a^{2}-\sqrt{2}a+12}{a+4}$$

Whakawehea ngā taha e rua ki te -1.

$$y=-\frac{-a^{2}-\sqrt{2}a+12}{a+4}$$

Kua oti te pūnaha te whakatau.

$$x=\frac{a+\sqrt{2}+3}{a+4},y=-\frac{-a^{2}-\sqrt{2}a+12}{a+4}$$