Whakaoti i te {l}{ax-by+8=0}{bx+ay+1=0}

$$\left\{ \begin{array} { l } { a x - b y + 8 = 0 } \\ { b x + a y + 1 = 0 } \end{array} \right.$$

Whakaoti mō x, y

$$x=-\frac{8a+b}{a^{2}+b^{2}}$$

$$y=-\frac{a-8b}{a^{2}+b^{2}}$$

$$b\neq 0\text{ or }a\neq 0$$

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.

$$ax+\left(-b\right)y+8=0,bx+ay+1=0$$

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

$$ax+\left(-b\right)y+8=0$$

Me tango 8 mai i ngā taha e rua o te whārite.

$$ax+\left(-b\right)y=-8$$

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

$$ax=by-8$$

Whakawehea ngā taha e rua ki te a.

$$x=\frac{1}{a}\left(by-8\right)$$

Whakareatia \frac{1}{a} ki te by-8.

$$x=\frac{b}{a}y-\frac{8}{a}$$

Whakakapia te \frac{by-8}{a} mō te x ki tērā atu whārite, bx+ay+1=0.

$$b\left(\frac{b}{a}y-\frac{8}{a}\right)+ay+1=0$$

Whakareatia b ki te \frac{by-8}{a}.

$$\frac{b^{2}}{a}y-\frac{8b}{a}+ay+1=0$$

Tāpiri \frac{b^{2}y}{a} ki te ay.

$$\left(\frac{b^{2}}{a}+a\right)y-\frac{8b}{a}+1=0$$

Tāpiri -\frac{8b}{a} ki te 1.

$$\left(\frac{b^{2}}{a}+a\right)y+\frac{a-8b}{a}=0$$

Me tango \frac{a-8b}{a} mai i ngā taha e rua o te whārite.

$$\left(\frac{b^{2}}{a}+a\right)y=\frac{8b}{a}-1$$

Whakawehea ngā taha e rua ki te a+\frac{b^{2}}{a}.

$$y=\frac{8b-a}{a^{2}+b^{2}}$$

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

$$x=\frac{b}{a}\times \frac{8b-a}{a^{2}+b^{2}}-\frac{8}{a}$$

Whakareatia \frac{b}{a} ki te \frac{8b-a}{a^{2}+b^{2}}.

$$x=\frac{b\left(8b-a\right)}{a\left(a^{2}+b^{2}\right)}-\frac{8}{a}$$

Tāpiri -\frac{8}{a} ki te \frac{b\left(8b-a\right)}{a\left(a^{2}+b^{2}\right)}.

$$x=-\frac{8a+b}{a^{2}+b^{2}}$$

Kua oti te pūnaha te whakatau.

$$x=-\frac{8a+b}{a^{2}+b^{2}},y=\frac{8b-a}{a^{2}+b^{2}}$$

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.

$$ax+\left(-b\right)y+8=0,bx+ay+1=0$$

Tuhia ngā whārite ki te tikanga tātai poukapa.

$$\left(\begin{matrix}a&-b\\b&a\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-8\\-1\end{matrix}\right)$$

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

$$inverse(\left(\begin{matrix}a&-b\\b&a\end{matrix}\right))\left(\begin{matrix}a&-b\\b&a\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}a&-b\\b&a\end{matrix}\right))\left(\begin{matrix}-8\\-1\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}a&-b\\b&a\end{matrix}\right))\left(\begin{matrix}-8\\-1\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}a&-b\\b&a\end{matrix}\right))\left(\begin{matrix}-8\\-1\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{a}{aa-\left(-b\right)b}&-\frac{-b}{aa-\left(-b\right)b}\\-\frac{b}{aa-\left(-b\right)b}&\frac{a}{aa-\left(-b\right)b}\end{matrix}\right)\left(\begin{matrix}-8\\-1\end{matrix}\right)$$

Mahia ngā tātaitanga.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{a}{a^{2}+b^{2}}&\frac{b}{a^{2}+b^{2}}\\-\frac{b}{a^{2}+b^{2}}&\frac{a}{a^{2}+b^{2}}\end{matrix}\right)\left(\begin{matrix}-8\\-1\end{matrix}\right)$$

Whakareatia ngā poukapa.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{a}{a^{2}+b^{2}}\left(-8\right)+\frac{b}{a^{2}+b^{2}}\left(-1\right)\\\left(-\frac{b}{a^{2}+b^{2}}\right)\left(-8\right)+\frac{a}{a^{2}+b^{2}}\left(-1\right)\end{matrix}\right)$$

Mahia ngā tātaitanga.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{8a+b}{a^{2}+b^{2}}\\\frac{8b-a}{a^{2}+b^{2}}\end{matrix}\right)$$

Tangohia ngā huānga poukapa x me y.

$$x=-\frac{8a+b}{a^{2}+b^{2}},y=\frac{8b-a}{a^{2}+b^{2}}$$

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.

$$ax+\left(-b\right)y+8=0,bx+ay+1=0$$

Kia ōrite ai a ax me bx, whakareatia ngā kīanga tau katoa kei ia taha o te whārite tuatahi ki te b me ngā kīanga tau katoa kei ia taha o te whārite tuarua ki te a.

$$bax+b\left(-b\right)y+b\times 8=0,abx+aay+a=0$$

Whakarūnātia.

$$abx+\left(-b^{2}\right)y+8b=0,abx+a^{2}y+a=0$$

Me tango abx+a^{2}y+a=0 mai i abx+\left(-b^{2}\right)y+8b=0 mā te tango i ngā kīanga tau ōrite i ia taha o te tohu ōrite.

$$abx+\left(-ab\right)x+\left(-b^{2}\right)y+\left(-a^{2}\right)y+8b-a=0$$

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

$$\left(-b^{2}\right)y+\left(-a^{2}\right)y+8b-a=0$$

Tāpiri -b^{2}y ki te -a^{2}y.

$$\left(-a^{2}-b^{2}\right)y+8b-a=0$$

Me tango 8b-a mai i ngā taha e rua o te whārite.

$$\left(-a^{2}-b^{2}\right)y=a-8b$$

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

$$y=-\frac{a-8b}{a^{2}+b^{2}}$$

Whakaurua te -\frac{-8b+a}{b^{2}+a^{2}} mō y ki bx+ay+1=0. I te mea kotahi anake te taurangi kei te whārite i puta, ka taea e koe te whakaoti mō x hāngai tonu.

$$bx+a\left(-\frac{a-8b}{a^{2}+b^{2}}\right)+1=0$$

Whakareatia a ki te -\frac{-8b+a}{b^{2}+a^{2}}.

$$bx-\frac{a\left(a-8b\right)}{a^{2}+b^{2}}+1=0$$

Tāpiri -\frac{a\left(-8b+a\right)}{b^{2}+a^{2}} ki te 1.

$$bx+\frac{b\left(8a+b\right)}{a^{2}+b^{2}}=0$$

Me tango \frac{b\left(8a+b\right)}{b^{2}+a^{2}} mai i ngā taha e rua o te whārite.

$$bx=-\frac{b\left(8a+b\right)}{a^{2}+b^{2}}$$

Whakawehea ngā taha e rua ki te b.

$$x=-\frac{8a+b}{a^{2}+b^{2}}$$

Kua oti te pūnaha te whakatau.

$$x=-\frac{8a+b}{a^{2}+b^{2}},y=-\frac{a-8b}{a^{2}+b^{2}}$$