Whakaoti i te {l}{Ax+By=C}{Dx+Cy=F}

$$\left. \begin{array} { l } { A x + B y = C } \\ { D x + C y = F } \end{array} \right.$$

Whakaoti mō x, y (complex solution)

$$\left\{\begin{matrix}x=-\frac{BF-C^{2}}{AC-BD}\text{, }y=-\frac{CD-AF}{AC-BD}\text{, }&\left(B\neq 0\text{ or }C\neq 0\right)\text{ and }\left(C\neq 0\text{ or }D\neq 0\right)\text{ and }\left(C=0\text{ or }A\neq \frac{BD}{C}\text{ or }B=0\text{ or }D=0\right)\text{ and }A\neq 0\\x=-\frac{By-C}{A}\text{, }y\in \mathrm{C}\text{, }&A\neq 0\text{ and }F=\frac{BD^{2}}{A^{2}}\text{ and }C=\frac{BD}{A}\\x=\frac{BF-C^{2}}{BD}\text{, }y=\frac{C}{B}\text{, }&A=0\text{ and }D\neq 0\text{ and }B\neq 0\\x=\frac{F}{D}\text{, }y\in \mathrm{C}\text{, }&A=0\text{ and }D\neq 0\text{ and }C=0\text{ and }B=0\\x\in \mathrm{C}\text{, }y=B^{-\frac{1}{2}}\sqrt{F}\text{, }&A=0\text{ and }D=0\text{ and }B\neq 0\text{ and }C=\sqrt{B}\sqrt{F}\\x\in \mathrm{C}\text{, }y=-B^{-\frac{1}{2}}\sqrt{F}\text{, }&A=0\text{ and }D=0\text{ and }B\neq 0\text{ and }C=-\sqrt{B}\sqrt{F}\\x\in \mathrm{C}\text{, }y\in \mathrm{C}\text{, }&A=0\text{ and }D=0\text{ and }F=0\text{ and }B=0\text{ and }C=0\end{matrix}\right.$$

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.

$$Ax+By=C,Dx+Cy=F$$

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, Dx+Cy=F.

$$D\left(\left(-\frac{B}{A}\right)y+\frac{C}{A}\right)+Cy=F$$

Whakareatia D ki te \frac{-By+C}{A}.

$$\left(-\frac{BD}{A}\right)y+\frac{CD}{A}+Cy=F$$

Tāpiri -\frac{DBy}{A} ki te Cy.

$$\left(-\frac{BD}{A}+C\right)y+\frac{CD}{A}=F$$

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

$$\left(-\frac{BD}{A}+C\right)y=-\frac{CD}{A}+F$$

Whakawehea ngā taha e rua ki te C-\frac{DB}{A}.

$$y=\frac{AF-CD}{AC-BD}$$

Whakaurua te \frac{FA-DC}{CA-DB} 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.

$$x=\left(-\frac{B}{A}\right)\times \frac{AF-CD}{AC-BD}+\frac{C}{A}$$

Whakareatia -\frac{B}{A} ki te \frac{FA-DC}{CA-DB}.

$$x=-\frac{B\left(AF-CD\right)}{A\left(AC-BD\right)}+\frac{C}{A}$$

Tāpiri \frac{C}{A} ki te -\frac{B\left(FA-DC\right)}{A\left(CA-DB\right)}.

$$x=\frac{C^{2}-BF}{AC-BD}$$

Kua oti te pūnaha te whakatau.

$$x=\frac{C^{2}-BF}{AC-BD},y=\frac{AF-CD}{AC-BD}$$

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.

$$Ax+By=C,Dx+Cy=F$$

Tuhia ngā whārite ki te tikanga tātai poukapa.

$$\left(\begin{matrix}A&B\\D&C\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}C\\F\end{matrix}\right)$$

Whakarea mauī i te whārite ki te poukapa kōaro o \left(\begin{matrix}A&B\\D&C\end{matrix}\right).

$$inverse(\left(\begin{matrix}A&B\\D&C\end{matrix}\right))\left(\begin{matrix}A&B\\D&C\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}A&B\\D&C\end{matrix}\right))\left(\begin{matrix}C\\F\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}A&B\\D&C\end{matrix}\right))\left(\begin{matrix}C\\F\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}A&B\\D&C\end{matrix}\right))\left(\begin{matrix}C\\F\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{C}{AC-BD}&-\frac{B}{AC-BD}\\-\frac{D}{AC-BD}&\frac{A}{AC-BD}\end{matrix}\right)\left(\begin{matrix}C\\F\end{matrix}\right)$$

Whakareatia ngā poukapa.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{C}{AC-BD}C+\left(-\frac{B}{AC-BD}\right)F\\\left(-\frac{D}{AC-BD}\right)C+\frac{A}{AC-BD}F\end{matrix}\right)$$

Mahia ngā tātaitanga.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{BF-C^{2}}{BD-AC}\\\frac{CD-AF}{BD-AC}\end{matrix}\right)$$

Tangohia ngā huānga poukapa x me y.

$$x=\frac{BF-C^{2}}{BD-AC},y=\frac{CD-AF}{BD-AC}$$

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.

$$Ax+By=C,Dx+Cy=F$$

Kia ōrite ai a Ax me Dx, whakareatia ngā kīanga tau katoa kei ia taha o te whārite tuatahi ki te D me ngā kīanga tau katoa kei ia taha o te whārite tuarua ki te A.

$$DAx+DBy=DC,ADx+ACy=AF$$

Whakarūnātia.

$$ADx+BDy=CD,ADx+ACy=AF$$

Me tango ADx+ACy=AF mai i ADx+BDy=CD mā te tango i ngā kīanga tau ōrite i ia taha o te tohu ōrite.

$$ADx+\left(-AD\right)x+BDy+\left(-AC\right)y=CD-AF$$

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

$$BDy+\left(-AC\right)y=CD-AF$$

Tāpiri DBy ki te -ACy.

$$\left(BD-AC\right)y=CD-AF$$

Whakawehea ngā taha e rua ki te DB-AC.

$$y=\frac{CD-AF}{BD-AC}$$

Whakaurua te \frac{DC-AF}{DB-AC} mō y ki Dx+Cy=F. I te mea kotahi anake te taurangi kei te whārite i puta, ka taea e koe te whakaoti mō x hāngai tonu.

$$Dx+C\times \frac{CD-AF}{BD-AC}=F$$

Whakareatia C ki te \frac{DC-AF}{DB-AC}.

$$Dx+\frac{C\left(CD-AF\right)}{BD-AC}=F$$

Me tango \frac{C\left(DC-AF\right)}{DB-AC} mai i ngā taha e rua o te whārite.

$$Dx=\frac{D\left(BF-C^{2}\right)}{BD-AC}$$

Whakawehea ngā taha e rua ki te D.

$$x=\frac{BF-C^{2}}{BD-AC}$$

Kua oti te pūnaha te whakatau.

$$x=\frac{BF-C^{2}}{BD-AC},y=\frac{CD-AF}{BD-AC}$$