Whakaoti i te {l}{y=kx+2}{y=2x+k}

$$\left\{ \begin{array} { l } { y = k x + 2 } \\ { y = 2 x + k } \end{array} \right.$$

Whakaoti mō x, y (complex solution)

$$\left\{\begin{matrix}\\x=1\text{, }y=k+2\text{, }&\text{unconditionally}\\x=\frac{y-2}{2}\text{, }y\in \mathrm{C}\text{, }&k=2\end{matrix}\right.$$

Ngā Upane e Whakamahi ana i te Whakakapinga

Whakaarohia te whārite tuatahi. Tangohia te kx mai i ngā taha e rua.

$$y-kx=2$$

Whakaarohia te whārite tuarua. Tangohia te 2x mai i ngā taha e rua.

$$y-2x=k$$

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.

$$y+\left(-k\right)x=2,y-2x=k$$

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

$$y+\left(-k\right)x=2$$

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

$$y=kx+2$$

Whakakapia te kx+2 mō te y ki tērā atu whārite, y-2x=k.

$$kx+2-2x=k$$

Tāpiri kx ki te -2x.

$$\left(k-2\right)x+2=k$$

Me tango 2 mai i ngā taha e rua o te whārite.

$$\left(k-2\right)x=k-2$$

Whakawehea ngā taha e rua ki te k-2.

$$x=1$$

Whakaurua te 1 mō x ki y=kx+2. I te mea kotahi anake te taurangi kei te whārite i puta, ka taea e koe te whakaoti mō y hāngai tonu.

$$y=k+2$$

Kua oti te pūnaha te whakatau.

$$y=k+2,x=1$$

Ngā Upane e Whakamahi ana i Ngā Poukapa

Whakaarohia te whārite tuatahi. Tangohia te kx mai i ngā taha e rua.

$$y-kx=2$$

Whakaarohia te whārite tuarua. Tangohia te 2x mai i ngā taha e rua.

$$y-2x=k$$

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

$$y+\left(-k\right)x=2,y-2x=k$$

Tuhia ngā whārite ki te tikanga tātai poukapa.

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

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

$$inverse(\left(\begin{matrix}1&-k\\1&-2\end{matrix}\right))\left(\begin{matrix}1&-k\\1&-2\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}1&-k\\1&-2\end{matrix}\right))\left(\begin{matrix}2\\k\end{matrix}\right)$$

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

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

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}1&-k\\1&-2\end{matrix}\right))\left(\begin{matrix}2\\k\end{matrix}\right)$$

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.

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

Mahia ngā tātaitanga.

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

Whakareatia ngā poukapa.

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

Mahia ngā tātaitanga.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}k+2\\1\end{matrix}\right)$$

Tangohia ngā huānga poukapa y me x.

$$y=k+2,x=1$$

Ngā Upane e Whakamahi i te Tangohanga

Whakaarohia te whārite tuatahi. Tangohia te kx mai i ngā taha e rua.

$$y-kx=2$$

Whakaarohia te whārite tuarua. Tangohia te 2x mai i ngā taha e rua.

$$y-2x=k$$

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.

$$y+\left(-k\right)x=2,y-2x=k$$

Me tango y-2x=k mai i y+\left(-k\right)x=2 mā te tango i ngā kīanga tau ōrite i ia taha o te tohu ōrite.

$$y-y+\left(-k\right)x+2x=2-k$$

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

$$\left(-k\right)x+2x=2-k$$

Tāpiri -kx ki te 2x.

$$\left(2-k\right)x=2-k$$

Whakawehea ngā taha e rua ki te -k+2.

$$x=1$$

Whakaurua te 1 mō x ki y-2x=k. I te mea kotahi anake te taurangi kei te whārite i puta, ka taea e koe te whakaoti mō y hāngai tonu.

$$y-2=k$$

Me tāpiri 2 ki ngā taha e rua o te whārite.

$$y=k+2$$

Kua oti te pūnaha te whakatau.

$$y=k+2,x=1$$

Whakaoti mō x, y

$$\left\{\begin{matrix}\\x=1\text{, }y=k+2\text{, }&\text{unconditionally}\\x=\frac{y-2}{2}\text{, }y\in \mathrm{R}\text{, }&k=2\end{matrix}\right.$$

Ngā Upane e Whakamahi ana i te Whakakapinga

Whakaarohia te whārite tuatahi. Tangohia te kx mai i ngā taha e rua.

$$y-kx=2$$

Whakaarohia te whārite tuarua. Tangohia te 2x mai i ngā taha e rua.

$$y-2x=k$$

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.

$$y+\left(-k\right)x=2,y-2x=k$$

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

$$y+\left(-k\right)x=2$$

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

$$y=kx+2$$

Whakakapia te kx+2 mō te y ki tērā atu whārite, y-2x=k.

$$kx+2-2x=k$$

Tāpiri kx ki te -2x.

$$\left(k-2\right)x+2=k$$

Me tango 2 mai i ngā taha e rua o te whārite.

$$\left(k-2\right)x=k-2$$

Whakawehea ngā taha e rua ki te k-2.

$$x=1$$

Whakaurua te 1 mō x ki y=kx+2. I te mea kotahi anake te taurangi kei te whārite i puta, ka taea e koe te whakaoti mō y hāngai tonu.

$$y=k+2$$

Kua oti te pūnaha te whakatau.

$$y=k+2,x=1$$

Ngā Upane e Whakamahi ana i Ngā Poukapa

Whakaarohia te whārite tuatahi. Tangohia te kx mai i ngā taha e rua.

$$y-kx=2$$

Whakaarohia te whārite tuarua. Tangohia te 2x mai i ngā taha e rua.

$$y-2x=k$$

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

$$y+\left(-k\right)x=2,y-2x=k$$

Tuhia ngā whārite ki te tikanga tātai poukapa.

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

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

$$inverse(\left(\begin{matrix}1&-k\\1&-2\end{matrix}\right))\left(\begin{matrix}1&-k\\1&-2\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}1&-k\\1&-2\end{matrix}\right))\left(\begin{matrix}2\\k\end{matrix}\right)$$

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

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

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}1&-k\\1&-2\end{matrix}\right))\left(\begin{matrix}2\\k\end{matrix}\right)$$

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.

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

Mahia ngā tātaitanga.

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

Whakareatia ngā poukapa.

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

Mahia ngā tātaitanga.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}k+2\\1\end{matrix}\right)$$

Tangohia ngā huānga poukapa y me x.

$$y=k+2,x=1$$

Ngā Upane e Whakamahi i te Tangohanga

Whakaarohia te whārite tuatahi. Tangohia te kx mai i ngā taha e rua.

$$y-kx=2$$

Whakaarohia te whārite tuarua. Tangohia te 2x mai i ngā taha e rua.

$$y-2x=k$$

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.

$$y+\left(-k\right)x=2,y-2x=k$$

Me tango y-2x=k mai i y+\left(-k\right)x=2 mā te tango i ngā kīanga tau ōrite i ia taha o te tohu ōrite.

$$y-y+\left(-k\right)x+2x=2-k$$

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

$$\left(-k\right)x+2x=2-k$$

Tāpiri -kx ki te 2x.

$$\left(2-k\right)x=2-k$$

Whakawehea ngā taha e rua ki te -k+2.

$$x=1$$

Whakaurua te 1 mō x ki y-2x=k. I te mea kotahi anake te taurangi kei te whārite i puta, ka taea e koe te whakaoti mō y hāngai tonu.

$$y-2=k$$

Me tāpiri 2 ki ngā taha e rua o te whārite.

$$y=k+2$$

Kua oti te pūnaha te whakatau.

$$y=k+2,x=1$$