Whakaoti i te {l}{x/4+y/3=0}{x/2+y/6=-frac{3}{2}}

$$\left. \begin{array} { l } { \frac { x } { 4 } + \frac { y } { 3 } = 0 } \\ { \frac { x } { 2 } + \frac { y } { 6 } = - \frac { 3 } { 2 } } \end{array} \right.$$

Whakaoti mō x, y

$$x=-4$$

$$y=3$$

Ngā Upane e Whakamahi ana i te Whakakapinga

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.

$$\frac{1}{4}x+\frac{1}{3}y=0,\frac{1}{2}x+\frac{1}{6}y=-\frac{3}{2}$$

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

$$\frac{1}{4}x+\frac{1}{3}y=0$$

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

$$\frac{1}{4}x=-\frac{1}{3}y$$

Me whakarea ngā taha e rua ki te 4.

$$x=4\left(-\frac{1}{3}\right)y$$

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

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

Whakakapia te -\frac{4y}{3} mō te x ki tērā atu whārite, \frac{1}{2}x+\frac{1}{6}y=-\frac{3}{2}.

$$\frac{1}{2}\left(-\frac{4}{3}\right)y+\frac{1}{6}y=-\frac{3}{2}$$

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

$$-\frac{2}{3}y+\frac{1}{6}y=-\frac{3}{2}$$

Tāpiri -\frac{2y}{3} ki te \frac{y}{6}.

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

Me whakarea ngā taha e rua ki te -2.

$$y=3$$

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

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

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

$$x=-4$$

Kua oti te pūnaha te whakatau.

$$x=-4,y=3$$

Ngā Upane e Whakamahi ana i Ngā Poukapa

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

$$\frac{1}{4}x+\frac{1}{3}y=0,\frac{1}{2}x+\frac{1}{6}y=-\frac{3}{2}$$

Tuhia ngā whārite ki te tikanga tātai poukapa.

$$\left(\begin{matrix}\frac{1}{4}&\frac{1}{3}\\\frac{1}{2}&\frac{1}{6}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\-\frac{3}{2}\end{matrix}\right)$$

Whakarea mauī i te whārite ki te poukapa kōaro o \left(\begin{matrix}\frac{1}{4}&\frac{1}{3}\\\frac{1}{2}&\frac{1}{6}\end{matrix}\right).

$$inverse(\left(\begin{matrix}\frac{1}{4}&\frac{1}{3}\\\frac{1}{2}&\frac{1}{6}\end{matrix}\right))\left(\begin{matrix}\frac{1}{4}&\frac{1}{3}\\\frac{1}{2}&\frac{1}{6}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{4}&\frac{1}{3}\\\frac{1}{2}&\frac{1}{6}\end{matrix}\right))\left(\begin{matrix}0\\-\frac{3}{2}\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{4}&\frac{1}{3}\\\frac{1}{2}&\frac{1}{6}\end{matrix}\right))\left(\begin{matrix}0\\-\frac{3}{2}\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{4}&\frac{1}{3}\\\frac{1}{2}&\frac{1}{6}\end{matrix}\right))\left(\begin{matrix}0\\-\frac{3}{2}\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{\frac{1}{6}}{\frac{1}{4}\times \frac{1}{6}-\frac{1}{3}\times \frac{1}{2}}&-\frac{\frac{1}{3}}{\frac{1}{4}\times \frac{1}{6}-\frac{1}{3}\times \frac{1}{2}}\\-\frac{\frac{1}{2}}{\frac{1}{4}\times \frac{1}{6}-\frac{1}{3}\times \frac{1}{2}}&\frac{\frac{1}{4}}{\frac{1}{4}\times \frac{1}{6}-\frac{1}{3}\times \frac{1}{2}}\end{matrix}\right)\left(\begin{matrix}0\\-\frac{3}{2}\end{matrix}\right)$$

Mahia ngā tātaitanga.

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

Whakareatia ngā poukapa.

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

Mahia ngā tātaitanga.

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

Tangohia ngā huānga poukapa x me y.

$$x=-4,y=3$$

Ngā Upane e Whakamahi i te Tangohanga

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.

$$\frac{1}{4}x+\frac{1}{3}y=0,\frac{1}{2}x+\frac{1}{6}y=-\frac{3}{2}$$

Kia ōrite ai a \frac{x}{4} 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 \frac{1}{4}.

$$\frac{1}{2}\times \frac{1}{4}x+\frac{1}{2}\times \frac{1}{3}y=0,\frac{1}{4}\times \frac{1}{2}x+\frac{1}{4}\times \frac{1}{6}y=\frac{1}{4}\left(-\frac{3}{2}\right)$$

Whakarūnātia.

$$\frac{1}{8}x+\frac{1}{6}y=0,\frac{1}{8}x+\frac{1}{24}y=-\frac{3}{8}$$

Me tango \frac{1}{8}x+\frac{1}{24}y=-\frac{3}{8} mai i \frac{1}{8}x+\frac{1}{6}y=0 mā te tango i ngā kīanga tau ōrite i ia taha o te tohu ōrite.

$$\frac{1}{8}x-\frac{1}{8}x+\frac{1}{6}y-\frac{1}{24}y=\frac{3}{8}$$

Tāpiri \frac{x}{8} ki te -\frac{x}{8}. Ka whakakore atu ngā kupu \frac{x}{8} me -\frac{x}{8}, ka toe he whārite me tētahi taurangi kotahi ka taea te whakaoti.

$$\frac{1}{6}y-\frac{1}{24}y=\frac{3}{8}$$

Tāpiri \frac{y}{6} ki te -\frac{y}{24}.

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

Me whakarea ngā taha e rua ki te 8.

$$y=3$$

Whakaurua te 3 mō y ki \frac{1}{2}x+\frac{1}{6}y=-\frac{3}{2}. I te mea kotahi anake te taurangi kei te whārite i puta, ka taea e koe te whakaoti mō x hāngai tonu.

$$\frac{1}{2}x+\frac{1}{6}\times 3=-\frac{3}{2}$$

Whakareatia \frac{1}{6} ki te 3.

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

Me tango \frac{1}{2} mai i ngā taha e rua o te whārite.

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

Me whakarea ngā taha e rua ki te 2.

$$x=-4$$

Kua oti te pūnaha te whakatau.

$$x=-4,y=3$$