Whakaoti i te 2x-5y=-3,y=-2x+1

$$2 x - 5 y = - 3 , y = - 2 x + 1$$

Whakaoti mō x, y

$$x=\frac{1}{6}\approx 0.166666667$$

$$y=\frac{2}{3}\approx 0.666666667$$

Ngā Upane e Whakamahi ana i te Whakakapinga

Whakaarohia te whārite tuarua. Me tāpiri te 2x ki ngā taha e rua.

$$y+2x=1$$

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.

$$2x-5y=-3,2x+y=1$$

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

$$2x-5y=-3$$

Me tāpiri 5y ki ngā taha e rua o te whārite.

$$2x=5y-3$$

Whakawehea ngā taha e rua ki te 2.

$$x=\frac{1}{2}\left(5y-3\right)$$

Whakareatia \frac{1}{2} ki te 5y-3.

$$x=\frac{5}{2}y-\frac{3}{2}$$

Whakakapia te \frac{5y-3}{2} mō te x ki tērā atu whārite, 2x+y=1.

$$2\left(\frac{5}{2}y-\frac{3}{2}\right)+y=1$$

Whakareatia 2 ki te \frac{5y-3}{2}.

$$5y-3+y=1$$

Tāpiri 5y ki te y.

$$6y-3=1$$

Me tāpiri 3 ki ngā taha e rua o te whārite.

$$6y=4$$

Whakawehea ngā taha e rua ki te 6.

$$y=\frac{2}{3}$$

Whakaurua te \frac{2}{3} mō y ki x=\frac{5}{2}y-\frac{3}{2}. I te mea kotahi anake te taurangi kei te whārite i puta, ka taea e koe te whakaoti mō x hāngai tonu.

$$x=\frac{5}{2}\times \frac{2}{3}-\frac{3}{2}$$

Whakareatia \frac{5}{2} ki te \frac{2}{3} mā te whakarea taurunga ki te taurunga me te tauraro ki te tauraro, ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$x=\frac{5}{3}-\frac{3}{2}$$

Tāpiri -\frac{3}{2} ki te \frac{5}{3} mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$x=\frac{1}{6}$$

Kua oti te pūnaha te whakatau.

$$x=\frac{1}{6},y=\frac{2}{3}$$

Ngā Upane e Whakamahi ana i Ngā Poukapa

Whakaarohia te whārite tuarua. Me tāpiri te 2x ki ngā taha e rua.

$$y+2x=1$$

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

$$2x-5y=-3,2x+y=1$$

Tuhia ngā whārite ki te tikanga tātai poukapa.

$$\left(\begin{matrix}2&-5\\2&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-3\\1\end{matrix}\right)$$

Whakarea mauī i te whārite ki te poukapa kōaro o \left(\begin{matrix}2&-5\\2&1\end{matrix}\right).

$$inverse(\left(\begin{matrix}2&-5\\2&1\end{matrix}\right))\left(\begin{matrix}2&-5\\2&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-5\\2&1\end{matrix}\right))\left(\begin{matrix}-3\\1\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-5\\2&1\end{matrix}\right))\left(\begin{matrix}-3\\1\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-5\\2&1\end{matrix}\right))\left(\begin{matrix}-3\\1\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2-\left(-5\times 2\right)}&-\frac{-5}{2-\left(-5\times 2\right)}\\-\frac{2}{2-\left(-5\times 2\right)}&\frac{2}{2-\left(-5\times 2\right)}\end{matrix}\right)\left(\begin{matrix}-3\\1\end{matrix}\right)$$

Mahia ngā tātaitanga.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{12}&\frac{5}{12}\\-\frac{1}{6}&\frac{1}{6}\end{matrix}\right)\left(\begin{matrix}-3\\1\end{matrix}\right)$$

Whakareatia ngā poukapa.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{12}\left(-3\right)+\frac{5}{12}\\-\frac{1}{6}\left(-3\right)+\frac{1}{6}\end{matrix}\right)$$

Mahia ngā tātaitanga.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{6}\\\frac{2}{3}\end{matrix}\right)$$

Tangohia ngā huānga poukapa x me y.

$$x=\frac{1}{6},y=\frac{2}{3}$$

Ngā Upane e Whakamahi i te Tangohanga

Whakaarohia te whārite tuarua. Me tāpiri te 2x ki ngā taha e rua.

$$y+2x=1$$

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.

$$2x-5y=-3,2x+y=1$$

Me tango 2x+y=1 mai i 2x-5y=-3 mā te tango i ngā kīanga tau ōrite i ia taha o te tohu ōrite.

$$2x-2x-5y-y=-3-1$$

Tāpiri 2x ki te -2x. Ka whakakore atu ngā kupu 2x me -2x, ka toe he whārite me tētahi taurangi kotahi ka taea te whakaoti.

$$-5y-y=-3-1$$

Tāpiri -5y ki te -y.

$$-6y=-3-1$$

Tāpiri -3 ki te -1.

$$-6y=-4$$

Whakawehea ngā taha e rua ki te -6.

$$y=\frac{2}{3}$$

Whakaurua te \frac{2}{3} mō y ki 2x+y=1. I te mea kotahi anake te taurangi kei te whārite i puta, ka taea e koe te whakaoti mō x hāngai tonu.

$$2x+\frac{2}{3}=1$$

Me tango \frac{2}{3} mai i ngā taha e rua o te whārite.

$$2x=\frac{1}{3}$$

Whakawehea ngā taha e rua ki te 2.

$$x=\frac{1}{6}$$

Kua oti te pūnaha te whakatau.

$$x=\frac{1}{6},y=\frac{2}{3}$$