Whakaoti i te {l}{ax+by=c}{a^2x+b^2y=c}

$$\left. \begin{array} { l } { a x + b y = c } \\ { a ^ { 2 } x + b ^ { 2 } y = c } \end{array} \right.$$

Whakaoti mō x, y (complex solution)

$$\left\{\begin{matrix}x=-\frac{c\left(1-b\right)}{a\left(b-a\right)}\text{, }y=-\frac{c\left(a-1\right)}{b\left(b-a\right)}\text{, }&b\neq 0\text{ and }a\neq b\text{ and }a\neq 0\\x=-\frac{by-c}{a}\text{, }y\in \mathrm{C}\text{, }&\left(c=0\text{ and }b=0\text{ and }a\neq 0\right)\text{ or }\left(c=0\text{ and }a=b\text{ and }b\neq 0\right)\text{ or }\left(a=1\text{ and }b=1\right)\text{ or }\left(a=1\text{ and }b=0\right)\\x\in \mathrm{C}\text{, }y=0\text{, }&c=0\text{ and }a=0\\x\in \mathrm{C}\text{, }y=c\text{, }&b=1\text{ and }a=0\\x\in \mathrm{C}\text{, }y\in \mathrm{C}\text{, }&c=0\text{ and }b=0\text{ and }a=0\end{matrix}\right.$$

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.

$$ax+by=c,a^{2}x+b^{2}y=c$$

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

$$ax+by=c$$

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

$$ax=\left(-b\right)y+c$$

Whakawehea ngā taha e rua ki te a.

$$x=\frac{1}{a}\left(\left(-b\right)y+c\right)$$

Whakareatia \frac{1}{a} ki te -by+c.

$$x=\left(-\frac{b}{a}\right)y+\frac{c}{a}$$

Whakakapia te \frac{-by+c}{a} mō te x ki tērā atu whārite, a^{2}x+b^{2}y=c.

$$a^{2}\left(\left(-\frac{b}{a}\right)y+\frac{c}{a}\right)+b^{2}y=c$$

Whakareatia a^{2} ki te \frac{-by+c}{a}.

$$\left(-ab\right)y+ac+b^{2}y=c$$

Tāpiri -bay ki te b^{2}y.

$$b\left(b-a\right)y+ac=c$$

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

$$b\left(b-a\right)y=c-ac$$

Whakawehea ngā taha e rua ki te b\left(b-a\right).

$$y=\frac{c\left(1-a\right)}{b\left(b-a\right)}$$

Whakaurua te \frac{c\left(1-a\right)}{b\left(b-a\right)} mō y ki x=\left(-\frac{b}{a}\right)y+\frac{c}{a}. I te mea kotahi anake te taurangi kei te whārite i puta, ka taea e koe te whakaoti mō x hāngai tonu.

$$x=\left(-\frac{b}{a}\right)\times \frac{c\left(1-a\right)}{b\left(b-a\right)}+\frac{c}{a}$$

Whakareatia -\frac{b}{a} ki te \frac{c\left(1-a\right)}{b\left(b-a\right)}.

$$x=-\frac{c\left(1-a\right)}{a\left(b-a\right)}+\frac{c}{a}$$

Tāpiri \frac{c}{a} ki te -\frac{\left(1-a\right)c}{\left(b-a\right)a}.

$$x=\frac{c\left(b-1\right)}{a\left(b-a\right)}$$

Kua oti te pūnaha te whakatau.

$$x=\frac{c\left(b-1\right)}{a\left(b-a\right)},y=\frac{c\left(1-a\right)}{b\left(b-a\right)}$$

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.

$$ax+by=c,a^{2}x+b^{2}y=c$$

Tuhia ngā whārite ki te tikanga tātai poukapa.

$$\left(\begin{matrix}a&b\\a^{2}&b^{2}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}c\\c\end{matrix}\right)$$

Whakarea mauī i te whārite ki te poukapa kōaro o \left(\begin{matrix}a&b\\a^{2}&b^{2}\end{matrix}\right).

$$inverse(\left(\begin{matrix}a&b\\a^{2}&b^{2}\end{matrix}\right))\left(\begin{matrix}a&b\\a^{2}&b^{2}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}a&b\\a^{2}&b^{2}\end{matrix}\right))\left(\begin{matrix}c\\c\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}a&b\\a^{2}&b^{2}\end{matrix}\right))\left(\begin{matrix}c\\c\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}a&b\\a^{2}&b^{2}\end{matrix}\right))\left(\begin{matrix}c\\c\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{b^{2}}{ab^{2}-ba^{2}}&-\frac{b}{ab^{2}-ba^{2}}\\-\frac{a^{2}}{ab^{2}-ba^{2}}&\frac{a}{ab^{2}-ba^{2}}\end{matrix}\right)\left(\begin{matrix}c\\c\end{matrix}\right)$$

Mahia ngā tātaitanga.

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

Whakareatia ngā poukapa.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{b}{a\left(b-a\right)}c+\left(-\frac{1}{a\left(b-a\right)}\right)c\\\left(-\frac{a}{b\left(b-a\right)}\right)c+\frac{1}{b\left(b-a\right)}c\end{matrix}\right)$$

Mahia ngā tātaitanga.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{c\left(b-1\right)}{a\left(b-a\right)}\\\frac{c\left(1-a\right)}{b\left(b-a\right)}\end{matrix}\right)$$

Tangohia ngā huānga poukapa x me y.

$$x=\frac{c\left(b-1\right)}{a\left(b-a\right)},y=\frac{c\left(1-a\right)}{b\left(b-a\right)}$$

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.

$$ax+by=c,a^{2}x+b^{2}y=c$$

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

$$a^{2}ax+a^{2}by=a^{2}c,aa^{2}x+ab^{2}y=ac$$

Whakarūnātia.

$$a^{3}x+ba^{2}y=ca^{2},a^{3}x+ab^{2}y=ac$$

Me tango a^{3}x+ab^{2}y=ac mai i a^{3}x+ba^{2}y=ca^{2} mā te tango i ngā kīanga tau ōrite i ia taha o te tohu ōrite.

$$a^{3}x+\left(-a^{3}\right)x+ba^{2}y+\left(-ab^{2}\right)y=ca^{2}-ac$$

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

$$ba^{2}y+\left(-ab^{2}\right)y=ca^{2}-ac$$

Tāpiri a^{2}by ki te -ab^{2}y.

$$ab\left(a-b\right)y=ca^{2}-ac$$

Tāpiri a^{2}c ki te -ac.

$$ab\left(a-b\right)y=ac\left(a-1\right)$$

Whakawehea ngā taha e rua ki te ab\left(a-b\right).

$$y=\frac{c\left(a-1\right)}{b\left(a-b\right)}$$

Whakaurua te \frac{\left(-1+a\right)c}{b\left(a-b\right)} mō y ki a^{2}x+b^{2}y=c. I te mea kotahi anake te taurangi kei te whārite i puta, ka taea e koe te whakaoti mō x hāngai tonu.

$$a^{2}x+b^{2}\times \frac{c\left(a-1\right)}{b\left(a-b\right)}=c$$

Whakareatia b^{2} ki te \frac{\left(-1+a\right)c}{b\left(a-b\right)}.

$$a^{2}x+\frac{bc\left(a-1\right)}{a-b}=c$$

Me tango \frac{b\left(-1+a\right)c}{a-b} mai i ngā taha e rua o te whārite.

$$a^{2}x=\frac{ac\left(1-b\right)}{a-b}$$

Whakawehea ngā taha e rua ki te a^{2}.

$$x=\frac{c\left(1-b\right)}{a\left(a-b\right)}$$

Kua oti te pūnaha te whakatau.

$$x=\frac{c\left(1-b\right)}{a\left(a-b\right)},y=\frac{c\left(a-1\right)}{b\left(a-b\right)}$$

Whakaoti mō x, y

$$\left\{\begin{matrix}x=-\frac{c\left(1-b\right)}{a\left(b-a\right)}\text{, }y=-\frac{c\left(a-1\right)}{b\left(b-a\right)}\text{, }&b\neq 0\text{ and }a\neq b\text{ and }a\neq 0\\x=-\frac{by-c}{a}\text{, }y\in \mathrm{R}\text{, }&\left(c=0\text{ and }b=0\text{ and }a\neq 0\right)\text{ or }\left(c=0\text{ and }a=b\text{ and }b\neq 0\right)\text{ or }\left(a=1\text{ and }b=1\right)\text{ or }\left(a=1\text{ and }b=0\right)\\x\in \mathrm{R}\text{, }y=0\text{, }&c=0\text{ and }a=0\text{ and }b\neq 1\text{ and }b\neq 0\\x\in \mathrm{R}\text{, }y=c\text{, }&b=1\text{ and }a=0\\x\in \mathrm{R}\text{, }y\in \mathrm{R}\text{, }&c=0\text{ and }b=0\text{ and }a=0\end{matrix}\right.$$

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.

$$ax+by=c,a^{2}x+b^{2}y=c$$

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

$$ax+by=c$$

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

$$ax=\left(-b\right)y+c$$

Whakawehea ngā taha e rua ki te a.

$$x=\frac{1}{a}\left(\left(-b\right)y+c\right)$$

Whakareatia \frac{1}{a} ki te -by+c.

$$x=\left(-\frac{b}{a}\right)y+\frac{c}{a}$$

Whakakapia te \frac{-by+c}{a} mō te x ki tērā atu whārite, a^{2}x+b^{2}y=c.

$$a^{2}\left(\left(-\frac{b}{a}\right)y+\frac{c}{a}\right)+b^{2}y=c$$

Whakareatia a^{2} ki te \frac{-by+c}{a}.

$$\left(-ab\right)y+ac+b^{2}y=c$$

Tāpiri -bay ki te b^{2}y.

$$b\left(b-a\right)y+ac=c$$

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

$$b\left(b-a\right)y=c-ac$$

Whakawehea ngā taha e rua ki te b\left(-a+b\right).

$$y=\frac{c\left(1-a\right)}{b\left(b-a\right)}$$

Whakaurua te \frac{c\left(1-a\right)}{b\left(-a+b\right)} mō y ki x=\left(-\frac{b}{a}\right)y+\frac{c}{a}. I te mea kotahi anake te taurangi kei te whārite i puta, ka taea e koe te whakaoti mō x hāngai tonu.

$$x=\left(-\frac{b}{a}\right)\times \frac{c\left(1-a\right)}{b\left(b-a\right)}+\frac{c}{a}$$

Whakareatia -\frac{b}{a} ki te \frac{c\left(1-a\right)}{b\left(-a+b\right)}.

$$x=-\frac{c\left(1-a\right)}{a\left(b-a\right)}+\frac{c}{a}$$

Tāpiri \frac{c}{a} ki te -\frac{\left(1-a\right)c}{\left(-a+b\right)a}.

$$x=\frac{c\left(b-1\right)}{a\left(b-a\right)}$$

Kua oti te pūnaha te whakatau.

$$x=\frac{c\left(b-1\right)}{a\left(b-a\right)},y=\frac{c\left(1-a\right)}{b\left(b-a\right)}$$

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.

$$ax+by=c,a^{2}x+b^{2}y=c$$

Tuhia ngā whārite ki te tikanga tātai poukapa.

$$\left(\begin{matrix}a&b\\a^{2}&b^{2}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}c\\c\end{matrix}\right)$$

Whakarea mauī i te whārite ki te poukapa kōaro o \left(\begin{matrix}a&b\\a^{2}&b^{2}\end{matrix}\right).

$$inverse(\left(\begin{matrix}a&b\\a^{2}&b^{2}\end{matrix}\right))\left(\begin{matrix}a&b\\a^{2}&b^{2}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}a&b\\a^{2}&b^{2}\end{matrix}\right))\left(\begin{matrix}c\\c\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}a&b\\a^{2}&b^{2}\end{matrix}\right))\left(\begin{matrix}c\\c\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}a&b\\a^{2}&b^{2}\end{matrix}\right))\left(\begin{matrix}c\\c\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{b^{2}}{ab^{2}-ba^{2}}&-\frac{b}{ab^{2}-ba^{2}}\\-\frac{a^{2}}{ab^{2}-ba^{2}}&\frac{a}{ab^{2}-ba^{2}}\end{matrix}\right)\left(\begin{matrix}c\\c\end{matrix}\right)$$

Mahia ngā tātaitanga.

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

Whakareatia ngā poukapa.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{b}{a\left(b-a\right)}c+\left(-\frac{1}{a\left(b-a\right)}\right)c\\\left(-\frac{a}{b\left(b-a\right)}\right)c+\frac{1}{b\left(b-a\right)}c\end{matrix}\right)$$

Mahia ngā tātaitanga.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{c\left(b-1\right)}{a\left(b-a\right)}\\\frac{c\left(1-a\right)}{b\left(b-a\right)}\end{matrix}\right)$$

Tangohia ngā huānga poukapa x me y.

$$x=\frac{c\left(b-1\right)}{a\left(b-a\right)},y=\frac{c\left(1-a\right)}{b\left(b-a\right)}$$

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.

$$ax+by=c,a^{2}x+b^{2}y=c$$

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

$$a^{2}ax+a^{2}by=a^{2}c,aa^{2}x+ab^{2}y=ac$$

Whakarūnātia.

$$a^{3}x+ba^{2}y=ca^{2},a^{3}x+ab^{2}y=ac$$

Me tango a^{3}x+ab^{2}y=ac mai i a^{3}x+ba^{2}y=ca^{2} mā te tango i ngā kīanga tau ōrite i ia taha o te tohu ōrite.

$$a^{3}x+\left(-a^{3}\right)x+ba^{2}y+\left(-ab^{2}\right)y=ca^{2}-ac$$

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

$$ba^{2}y+\left(-ab^{2}\right)y=ca^{2}-ac$$

Tāpiri a^{2}by ki te -ab^{2}y.

$$ab\left(a-b\right)y=ca^{2}-ac$$

Tāpiri a^{2}c ki te -ac.

$$ab\left(a-b\right)y=ac\left(a-1\right)$$

Whakawehea ngā taha e rua ki te ab\left(a-b\right).

$$y=\frac{c\left(a-1\right)}{b\left(a-b\right)}$$

Whakaurua te \frac{\left(-1+a\right)c}{b\left(a-b\right)} mō y ki a^{2}x+b^{2}y=c. I te mea kotahi anake te taurangi kei te whārite i puta, ka taea e koe te whakaoti mō x hāngai tonu.

$$a^{2}x+b^{2}\times \frac{c\left(a-1\right)}{b\left(a-b\right)}=c$$

Whakareatia b^{2} ki te \frac{\left(-1+a\right)c}{b\left(a-b\right)}.

$$a^{2}x+\frac{bc\left(a-1\right)}{a-b}=c$$

Me tango \frac{b\left(-1+a\right)c}{a-b} mai i ngā taha e rua o te whārite.

$$a^{2}x=\frac{ac\left(1-b\right)}{a-b}$$

Whakawehea ngā taha e rua ki te a^{2}.

$$x=\frac{c\left(1-b\right)}{a\left(a-b\right)}$$

Kua oti te pūnaha te whakatau.

$$x=\frac{c\left(1-b\right)}{a\left(a-b\right)},y=\frac{c\left(a-1\right)}{b\left(a-b\right)}$$