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

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

Whakaoti mō x, y

$$x=\frac{3}{4}=0.75$$

$$y = \frac{5}{4} = 1\frac{1}{4} = 1.25$$

Ngā Upane e Whakamahi ana i te Whakakapinga

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

$$x+\frac{1}{2}-y=0$$

Tangohia te \frac{1}{2} mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

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

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.

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

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

$$x+y=2$$

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

$$x=-y+2$$

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

$$-y+2-y=-\frac{1}{2}$$

Tāpiri -y ki te -y.

$$-2y+2=-\frac{1}{2}$$

Me tango 2 mai i ngā taha e rua o te whārite.

$$-2y=-\frac{5}{2}$$

Whakawehea ngā taha e rua ki te -2.

$$y=\frac{5}{4}$$

Whakaurua te \frac{5}{4} mō y ki x=-y+2. I te mea kotahi anake te taurangi kei te whārite i puta, ka taea e koe te whakaoti mō x hāngai tonu.

$$x=-\frac{5}{4}+2$$

Tāpiri 2 ki te -\frac{5}{4}.

$$x=\frac{3}{4}$$

Kua oti te pūnaha te whakatau.

$$x=\frac{3}{4},y=\frac{5}{4}$$

Ngā Upane e Whakamahi ana i Ngā Poukapa

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

$$x+\frac{1}{2}-y=0$$

Tangohia te \frac{1}{2} mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

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

Tuhia ngā whārite ki te tānga ngahuru ka whakamahi i ngā poukapa hei whakaoti i te pūnaha o ngā whārite.

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

Tuhia ngā whārite ki te tikanga tātai poukapa.

$$\left(\begin{matrix}1&1\\1&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2\\-\frac{1}{2}\end{matrix}\right)$$

Whakarea mauī i te whārite ki te poukapa kōaro o \left(\begin{matrix}1&1\\1&-1\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\1&-1\end{matrix}\right))\left(\begin{matrix}1&1\\1&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\1&-1\end{matrix}\right))\left(\begin{matrix}2\\-\frac{1}{2}\end{matrix}\right)$$

Ko te hua o tētahi poukapa me te kōaro ko te poukapa tuakiri.

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\1&-1\end{matrix}\right))\left(\begin{matrix}2\\-\frac{1}{2}\end{matrix}\right)$$

Whakareatia ngā poukapa kei te taha mauī o te tohu ōrite.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\1&-1\end{matrix}\right))\left(\begin{matrix}2\\-\frac{1}{2}\end{matrix}\right)$$

Mō te poukapa 2\times 2 \left(\begin{matrix}a&b\\c&d\end{matrix}\right), ko te \left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right) te poukapa kōaro, nō reira ka taea te tuhi anō te whārite poukapa hei rapanga whakarea poukapa.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{-1-1}&-\frac{1}{-1-1}\\-\frac{1}{-1-1}&\frac{1}{-1-1}\end{matrix}\right)\left(\begin{matrix}2\\-\frac{1}{2}\end{matrix}\right)$$

Mahia ngā tātaitanga.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2}&\frac{1}{2}\\\frac{1}{2}&-\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}2\\-\frac{1}{2}\end{matrix}\right)$$

Whakareatia ngā poukapa.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2}\times 2+\frac{1}{2}\left(-\frac{1}{2}\right)\\\frac{1}{2}\times 2-\frac{1}{2}\left(-\frac{1}{2}\right)\end{matrix}\right)$$

Mahia ngā tātaitanga.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{4}\\\frac{5}{4}\end{matrix}\right)$$

Tangohia ngā huānga poukapa x me y.

$$x=\frac{3}{4},y=\frac{5}{4}$$

Ngā Upane e Whakamahi i te Tangohanga

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

$$x+\frac{1}{2}-y=0$$

Tangohia te \frac{1}{2} mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

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

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.

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

Me tango x-y=-\frac{1}{2} mai i x+y=2 mā te tango i ngā kīanga tau ōrite i ia taha o te tohu ōrite.

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

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

$$y+y=2+\frac{1}{2}$$

Tāpiri y ki te y.

$$2y=2+\frac{1}{2}$$

Tāpiri 2 ki te \frac{1}{2}.

$$2y=\frac{5}{2}$$

Whakawehea ngā taha e rua ki te 2.

$$y=\frac{5}{4}$$

Whakaurua te \frac{5}{4} mō y ki x-y=-\frac{1}{2}. I te mea kotahi anake te taurangi kei te whārite i puta, ka taea e koe te whakaoti mō x hāngai tonu.

$$x-\frac{5}{4}=-\frac{1}{2}$$

Me tāpiri \frac{5}{4} ki ngā taha e rua o te whārite.

$$x=\frac{3}{4}$$

Kua oti te pūnaha te whakatau.

$$x=\frac{3}{4},y=\frac{5}{4}$$