Whakaoti i te {l}{5(4x-y)=0}{4(x+3)=13y}

$$\left. \begin{array} { l } { 5 ( 4 x - y ) = 0 } \\ { 4 ( x + 3 ) = 13 y } \end{array} \right.$$

Whakaoti mō x, y

$$x=\frac{1}{4}=0.25$$

$$y=1$$

Ngā Upane e Whakamahi ana i te Whakakapinga

Whakaarohia te whārite tuatahi. Whakawehea ngā taha e rua ki te 5. Ko te kore i whakawehea ki te tau ehara te kore ka hua ko te kore.

$$4x-y=0$$

Whakaarohia te whārite tuarua. Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te x+3.

$$4x+12=13y$$

Tangohia te 13y mai i ngā taha e rua.

$$4x+12-13y=0$$

Tangohia te 12 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$4x-13y=-12$$

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.

$$4x-y=0,4x-13y=-12$$

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

$$4x-y=0$$

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

$$4x=y$$

Whakawehea ngā taha e rua ki te 4.

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

Whakakapia te \frac{y}{4} mō te x ki tērā atu whārite, 4x-13y=-12.

$$4\times \frac{1}{4}y-13y=-12$$

Whakareatia 4 ki te \frac{y}{4}.

$$y-13y=-12$$

Tāpiri y ki te -13y.

$$-12y=-12$$

Whakawehea ngā taha e rua ki te -12.

$$y=1$$

Whakaurua te 1 mō y ki x=\frac{1}{4}y. I te mea kotahi anake te taurangi kei te whārite i puta, ka taea e koe te whakaoti mō x hāngai tonu.

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

Kua oti te pūnaha te whakatau.

$$x=\frac{1}{4},y=1$$

Ngā Upane e Whakamahi ana i Ngā Poukapa

Whakaarohia te whārite tuatahi. Whakawehea ngā taha e rua ki te 5. Ko te kore i whakawehea ki te tau ehara te kore ka hua ko te kore.

$$4x-y=0$$

Whakaarohia te whārite tuarua. Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te x+3.

$$4x+12=13y$$

Tangohia te 13y mai i ngā taha e rua.

$$4x+12-13y=0$$

Tangohia te 12 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$4x-13y=-12$$

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

$$4x-y=0,4x-13y=-12$$

Tuhia ngā whārite ki te tikanga tātai poukapa.

$$\left(\begin{matrix}4&-1\\4&-13\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\-12\end{matrix}\right)$$

Whakarea mauī i te whārite ki te poukapa kōaro o \left(\begin{matrix}4&-1\\4&-13\end{matrix}\right).

$$inverse(\left(\begin{matrix}4&-1\\4&-13\end{matrix}\right))\left(\begin{matrix}4&-1\\4&-13\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}4&-1\\4&-13\end{matrix}\right))\left(\begin{matrix}0\\-12\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}4&-1\\4&-13\end{matrix}\right))\left(\begin{matrix}0\\-12\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}4&-1\\4&-13\end{matrix}\right))\left(\begin{matrix}0\\-12\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{13}{4\left(-13\right)-\left(-4\right)}&-\frac{-1}{4\left(-13\right)-\left(-4\right)}\\-\frac{4}{4\left(-13\right)-\left(-4\right)}&\frac{4}{4\left(-13\right)-\left(-4\right)}\end{matrix}\right)\left(\begin{matrix}0\\-12\end{matrix}\right)$$

Mahia ngā tātaitanga.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{13}{48}&-\frac{1}{48}\\\frac{1}{12}&-\frac{1}{12}\end{matrix}\right)\left(\begin{matrix}0\\-12\end{matrix}\right)$$

Whakareatia ngā poukapa.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{48}\left(-12\right)\\-\frac{1}{12}\left(-12\right)\end{matrix}\right)$$

Mahia ngā tātaitanga.

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

Tangohia ngā huānga poukapa x me y.

$$x=\frac{1}{4},y=1$$

Ngā Upane e Whakamahi i te Tangohanga

Whakaarohia te whārite tuatahi. Whakawehea ngā taha e rua ki te 5. Ko te kore i whakawehea ki te tau ehara te kore ka hua ko te kore.

$$4x-y=0$$

Whakaarohia te whārite tuarua. Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te x+3.

$$4x+12=13y$$

Tangohia te 13y mai i ngā taha e rua.

$$4x+12-13y=0$$

Tangohia te 12 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$4x-13y=-12$$

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.

$$4x-y=0,4x-13y=-12$$

Me tango 4x-13y=-12 mai i 4x-y=0 mā te tango i ngā kīanga tau ōrite i ia taha o te tohu ōrite.

$$4x-4x-y+13y=12$$

Tāpiri 4x ki te -4x. Ka whakakore atu ngā kupu 4x me -4x, ka toe he whārite me tētahi taurangi kotahi ka taea te whakaoti.

$$-y+13y=12$$

Tāpiri -y ki te 13y.

$$12y=12$$

Whakawehea ngā taha e rua ki te 12.

$$y=1$$

Whakaurua te 1 mō y ki 4x-13y=-12. I te mea kotahi anake te taurangi kei te whārite i puta, ka taea e koe te whakaoti mō x hāngai tonu.

$$4x-13=-12$$

Me tāpiri 13 ki ngā taha e rua o te whārite.

$$4x=1$$

Whakawehea ngā taha e rua ki te 4.

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

Kua oti te pūnaha te whakatau.

$$x=\frac{1}{4},y=1$$