Whakaoti i te {l}{3/2x+1/3y=1}{x/4-1/6y=-frac{3}{2}}

$$\left\{ \begin{array} { l } { \frac { 3 } { 2 } x + \frac { 1 } { 3 } y = 1 } \\ { \frac { x } { 4 } - \frac { 1 } { 6 } y = - \frac { 3 } { 2 } } \end{array} \right.$$

Whakaoti mō x, y

$$x=-1$$

$$y = \frac{15}{2} = 7\frac{1}{2} = 7.5$$

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.

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

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

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

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

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

Whakawehea ngā taha e rua o te whārite ki te \frac{3}{2}, he ōrite ki te whakarea i ngā taha e rua ki te tau huripoki o te hautanga.

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

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

$$x=-\frac{2}{9}y+\frac{2}{3}$$

Whakakapia te -\frac{2y}{9}+\frac{2}{3} mō te x ki tērā atu whārite, \frac{1}{4}x-\frac{1}{6}y=-\frac{3}{2}.

$$\frac{1}{4}\left(-\frac{2}{9}y+\frac{2}{3}\right)-\frac{1}{6}y=-\frac{3}{2}$$

Whakareatia \frac{1}{4} ki te -\frac{2y}{9}+\frac{2}{3}.

$$-\frac{1}{18}y+\frac{1}{6}-\frac{1}{6}y=-\frac{3}{2}$$

Tāpiri -\frac{y}{18} ki te -\frac{y}{6}.

$$-\frac{2}{9}y+\frac{1}{6}=-\frac{3}{2}$$

Me tango \frac{1}{6} mai i ngā taha e rua o te whārite.

$$-\frac{2}{9}y=-\frac{5}{3}$$

Whakawehea ngā taha e rua o te whārite ki te -\frac{2}{9}, he ōrite ki te whakarea i ngā taha e rua ki te tau huripoki o te hautanga.

$$y=\frac{15}{2}$$

Whakaurua te \frac{15}{2} mō y ki x=-\frac{2}{9}y+\frac{2}{3}. I te mea kotahi anake te taurangi kei te whārite i puta, ka taea e koe te whakaoti mō x hāngai tonu.

$$x=-\frac{2}{9}\times \frac{15}{2}+\frac{2}{3}$$

Whakareatia -\frac{2}{9} ki te \frac{15}{2} mā te whakarea taurunga ki te taurunga me te tauraro ki te tauraro, ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$x=\frac{-5+2}{3}$$

Tāpiri \frac{2}{3} ki te -\frac{5}{3} mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$x=-1$$

Kua oti te pūnaha te whakatau.

$$x=-1,y=\frac{15}{2}$$

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.

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

Tuhia ngā whārite ki te tikanga tātai poukapa.

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

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

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

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

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

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

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

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.

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

Mahia ngā tātaitanga.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2}&1\\\frac{3}{4}&-\frac{9}{2}\end{matrix}\right)\left(\begin{matrix}1\\-\frac{3}{2}\end{matrix}\right)$$

Whakareatia ngā poukapa.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1-3}{2}\\\frac{3}{4}-\frac{9}{2}\left(-\frac{3}{2}\right)\end{matrix}\right)$$

Mahia ngā tātaitanga.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-1\\\frac{15}{2}\end{matrix}\right)$$

Tangohia ngā huānga poukapa x me y.

$$x=-1,y=\frac{15}{2}$$

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.

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

Kia ōrite ai a \frac{3x}{2} me \frac{x}{4}, whakareatia ngā kīanga tau katoa kei ia taha o te whārite tuatahi ki te \frac{1}{4} me ngā kīanga tau katoa kei ia taha o te whārite tuarua ki te \frac{3}{2}.

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

Whakarūnātia.

$$\frac{3}{8}x+\frac{1}{12}y=\frac{1}{4},\frac{3}{8}x-\frac{1}{4}y=-\frac{9}{4}$$

Me tango \frac{3}{8}x-\frac{1}{4}y=-\frac{9}{4} mai i \frac{3}{8}x+\frac{1}{12}y=\frac{1}{4} mā te tango i ngā kīanga tau ōrite i ia taha o te tohu ōrite.

$$\frac{3}{8}x-\frac{3}{8}x+\frac{1}{12}y+\frac{1}{4}y=\frac{1+9}{4}$$

Tāpiri \frac{3x}{8} ki te -\frac{3x}{8}. Ka whakakore atu ngā kupu \frac{3x}{8} me -\frac{3x}{8}, ka toe he whārite me tētahi taurangi kotahi ka taea te whakaoti.

$$\frac{1}{12}y+\frac{1}{4}y=\frac{1+9}{4}$$

Tāpiri \frac{y}{12} ki te \frac{y}{4}.

$$\frac{1}{3}y=\frac{1+9}{4}$$

Tāpiri \frac{1}{4} ki te \frac{9}{4} mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$\frac{1}{3}y=\frac{5}{2}$$

Me whakarea ngā taha e rua ki te 3.

$$y=\frac{15}{2}$$

Whakaurua te \frac{15}{2} mō y ki \frac{1}{4}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.

$$\frac{1}{4}x-\frac{1}{6}\times \frac{15}{2}=-\frac{3}{2}$$

Whakareatia -\frac{1}{6} ki te \frac{15}{2} mā te whakarea taurunga ki te taurunga me te tauraro ki te tauraro, ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$\frac{1}{4}x-\frac{5}{4}=-\frac{3}{2}$$

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

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

Me whakarea ngā taha e rua ki te 4.

$$x=-1$$

Kua oti te pūnaha te whakatau.

$$x=-1,y=\frac{15}{2}$$