Whakaoti i te {l}{x=3y+4}{y=1/2x-8/3}

$$\left\{ \begin{array} { l } { x = 3 y + 4 } \\ { y = \frac { 1 } { 2 } x - \frac { 8 } { 3 } } \end{array} \right.$$

Whakaoti mō x, y

$$x=8$$

$$y = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

Ngā Upane e Whakamahi ana i te Whakakapinga

Whakaarohia te whārite tuatahi. Tangohia te 3y mai i ngā taha e rua.

$$x-3y=4$$

Whakaarohia te whārite tuarua. Tangohia te \frac{1}{2}x mai i ngā taha e rua.

$$y-\frac{1}{2}x=-\frac{8}{3}$$

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.

$$x-3y=4,-\frac{1}{2}x+y=-\frac{8}{3}$$

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

$$x-3y=4$$

Me tāpiri 3y ki ngā taha e rua o te whārite.

$$x=3y+4$$

Whakakapia te 3y+4 mō te x ki tērā atu whārite, -\frac{1}{2}x+y=-\frac{8}{3}.

$$-\frac{1}{2}\left(3y+4\right)+y=-\frac{8}{3}$$

Whakareatia -\frac{1}{2} ki te 3y+4.

$$-\frac{3}{2}y-2+y=-\frac{8}{3}$$

Tāpiri -\frac{3y}{2} ki te y.

$$-\frac{1}{2}y-2=-\frac{8}{3}$$

Me tāpiri 2 ki ngā taha e rua o te whārite.

$$-\frac{1}{2}y=-\frac{2}{3}$$

Me whakarea ngā taha e rua ki te -2.

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

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

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

Whakareatia 3 ki te \frac{4}{3}.

$$x=4+4$$

Tāpiri 4 ki te 4.

$$x=8$$

Kua oti te pūnaha te whakatau.

$$x=8,y=\frac{4}{3}$$

Ngā Upane e Whakamahi ana i Ngā Poukapa

Whakaarohia te whārite tuatahi. Tangohia te 3y mai i ngā taha e rua.

$$x-3y=4$$

Whakaarohia te whārite tuarua. Tangohia te \frac{1}{2}x mai i ngā taha e rua.

$$y-\frac{1}{2}x=-\frac{8}{3}$$

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

$$x-3y=4,-\frac{1}{2}x+y=-\frac{8}{3}$$

Tuhia ngā whārite ki te tikanga tātai poukapa.

$$\left(\begin{matrix}1&-3\\-\frac{1}{2}&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}4\\-\frac{8}{3}\end{matrix}\right)$$

Whakarea mauī i te whārite ki te poukapa kōaro o \left(\begin{matrix}1&-3\\-\frac{1}{2}&1\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&-3\\-\frac{1}{2}&1\end{matrix}\right))\left(\begin{matrix}1&-3\\-\frac{1}{2}&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-3\\-\frac{1}{2}&1\end{matrix}\right))\left(\begin{matrix}4\\-\frac{8}{3}\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-3\\-\frac{1}{2}&1\end{matrix}\right))\left(\begin{matrix}4\\-\frac{8}{3}\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-3\\-\frac{1}{2}&1\end{matrix}\right))\left(\begin{matrix}4\\-\frac{8}{3}\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{1-\left(-3\left(-\frac{1}{2}\right)\right)}&-\frac{-3}{1-\left(-3\left(-\frac{1}{2}\right)\right)}\\-\frac{-\frac{1}{2}}{1-\left(-3\left(-\frac{1}{2}\right)\right)}&\frac{1}{1-\left(-3\left(-\frac{1}{2}\right)\right)}\end{matrix}\right)\left(\begin{matrix}4\\-\frac{8}{3}\end{matrix}\right)$$

Mahia ngā tātaitanga.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-2&-6\\-1&-2\end{matrix}\right)\left(\begin{matrix}4\\-\frac{8}{3}\end{matrix}\right)$$

Whakareatia ngā poukapa.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-2\times 4-6\left(-\frac{8}{3}\right)\\-4-2\left(-\frac{8}{3}\right)\end{matrix}\right)$$

Mahia ngā tātaitanga.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}8\\\frac{4}{3}\end{matrix}\right)$$

Tangohia ngā huānga poukapa x me y.

$$x=8,y=\frac{4}{3}$$

Ngā Upane e Whakamahi i te Tangohanga

Whakaarohia te whārite tuatahi. Tangohia te 3y mai i ngā taha e rua.

$$x-3y=4$$

Whakaarohia te whārite tuarua. Tangohia te \frac{1}{2}x mai i ngā taha e rua.

$$y-\frac{1}{2}x=-\frac{8}{3}$$

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.

$$x-3y=4,-\frac{1}{2}x+y=-\frac{8}{3}$$

Kia ōrite ai a x me -\frac{x}{2}, whakareatia ngā kīanga tau katoa kei ia taha o te whārite tuatahi ki te -\frac{1}{2} me ngā kīanga tau katoa kei ia taha o te whārite tuarua ki te 1.

$$-\frac{1}{2}x-\frac{1}{2}\left(-3\right)y=-\frac{1}{2}\times 4,-\frac{1}{2}x+y=-\frac{8}{3}$$

Whakarūnātia.

$$-\frac{1}{2}x+\frac{3}{2}y=-2,-\frac{1}{2}x+y=-\frac{8}{3}$$

Me tango -\frac{1}{2}x+y=-\frac{8}{3} mai i -\frac{1}{2}x+\frac{3}{2}y=-2 mā te tango i ngā kīanga tau ōrite i ia taha o te tohu ōrite.

$$-\frac{1}{2}x+\frac{1}{2}x+\frac{3}{2}y-y=-2+\frac{8}{3}$$

Tāpiri -\frac{x}{2} ki te \frac{x}{2}. Ka whakakore atu ngā kupu -\frac{x}{2} me \frac{x}{2}, ka toe he whārite me tētahi taurangi kotahi ka taea te whakaoti.

$$\frac{3}{2}y-y=-2+\frac{8}{3}$$

Tāpiri \frac{3y}{2} ki te -y.

$$\frac{1}{2}y=-2+\frac{8}{3}$$

Tāpiri -2 ki te \frac{8}{3}.

$$\frac{1}{2}y=\frac{2}{3}$$

Me whakarea ngā taha e rua ki te 2.

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

Whakaurua te \frac{4}{3} mō y ki -\frac{1}{2}x+y=-\frac{8}{3}. I te mea kotahi anake te taurangi kei te whārite i puta, ka taea e koe te whakaoti mō x hāngai tonu.

$$-\frac{1}{2}x+\frac{4}{3}=-\frac{8}{3}$$

Me tango \frac{4}{3} mai i ngā taha e rua o te whārite.

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

Me whakarea ngā taha e rua ki te -2.

$$x=8$$

Kua oti te pūnaha te whakatau.

$$x=8,y=\frac{4}{3}$$