Whakaoti i te {l}{0.5x-0.8y+9=4}{x/3+y/5=4}

$$\left\{ \begin{array} { l } { 0.5 x - 0.8 y + 9 = 4 } \\ { \frac { x } { 3 } + \frac { y } { 5 } = 4 } \end{array} \right.$$

Whakaoti mō x, y

$$x=6$$

$$y=10$$

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.

$$0.5x-0.8y+9=4,\frac{1}{3}x+\frac{1}{5}y=4$$

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

$$0.5x-0.8y+9=4$$

Me tango 9 mai i ngā taha e rua o te whārite.

$$0.5x-0.8y=-5$$

Me tāpiri \frac{4y}{5} ki ngā taha e rua o te whārite.

$$0.5x=0.8y-5$$

Me whakarea ngā taha e rua ki te 2.

$$x=2\left(0.8y-5\right)$$

Whakareatia 2 ki te \frac{4y}{5}-5.

$$x=1.6y-10$$

Whakakapia te \frac{8y}{5}-10 mō te x ki tērā atu whārite, \frac{1}{3}x+\frac{1}{5}y=4.

$$\frac{1}{3}\left(1.6y-10\right)+\frac{1}{5}y=4$$

Whakareatia \frac{1}{3} ki te \frac{8y}{5}-10.

$$\frac{8}{15}y-\frac{10}{3}+\frac{1}{5}y=4$$

Tāpiri \frac{8y}{15} ki te \frac{y}{5}.

$$\frac{11}{15}y-\frac{10}{3}=4$$

Me tāpiri \frac{10}{3} ki ngā taha e rua o te whārite.

$$\frac{11}{15}y=\frac{22}{3}$$

Whakawehea ngā taha e rua o te whārite ki te \frac{11}{15}, he ōrite ki te whakarea i ngā taha e rua ki te tau huripoki o te hautanga.

$$y=10$$

Whakaurua te 10 mō y ki x=1.6y-10. I te mea kotahi anake te taurangi kei te whārite i puta, ka taea e koe te whakaoti mō x hāngai tonu.

$$x=1.6\times 10-10$$

Whakareatia 1.6 ki te 10.

$$x=16-10$$

Tāpiri -10 ki te 16.

$$x=6$$

Kua oti te pūnaha te whakatau.

$$x=6,y=10$$

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.

$$0.5x-0.8y+9=4,\frac{1}{3}x+\frac{1}{5}y=4$$

Tuhia ngā whārite ki te tikanga tātai poukapa.

$$\left(\begin{matrix}0.5&-0.8\\\frac{1}{3}&\frac{1}{5}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-5\\4\end{matrix}\right)$$

Whakarea mauī i te whārite ki te poukapa kōaro o \left(\begin{matrix}0.5&-0.8\\\frac{1}{3}&\frac{1}{5}\end{matrix}\right).

$$inverse(\left(\begin{matrix}0.5&-0.8\\\frac{1}{3}&\frac{1}{5}\end{matrix}\right))\left(\begin{matrix}0.5&-0.8\\\frac{1}{3}&\frac{1}{5}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}0.5&-0.8\\\frac{1}{3}&\frac{1}{5}\end{matrix}\right))\left(\begin{matrix}-5\\4\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}0.5&-0.8\\\frac{1}{3}&\frac{1}{5}\end{matrix}\right))\left(\begin{matrix}-5\\4\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}0.5&-0.8\\\frac{1}{3}&\frac{1}{5}\end{matrix}\right))\left(\begin{matrix}-5\\4\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{\frac{1}{5}}{0.5\times \frac{1}{5}-\left(-0.8\times \frac{1}{3}\right)}&-\frac{-0.8}{0.5\times \frac{1}{5}-\left(-0.8\times \frac{1}{3}\right)}\\-\frac{\frac{1}{3}}{0.5\times \frac{1}{5}-\left(-0.8\times \frac{1}{3}\right)}&\frac{0.5}{0.5\times \frac{1}{5}-\left(-0.8\times \frac{1}{3}\right)}\end{matrix}\right)\left(\begin{matrix}-5\\4\end{matrix}\right)$$

Mahia ngā tātaitanga.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{6}{11}&\frac{24}{11}\\-\frac{10}{11}&\frac{15}{11}\end{matrix}\right)\left(\begin{matrix}-5\\4\end{matrix}\right)$$

Whakareatia ngā poukapa.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{6}{11}\left(-5\right)+\frac{24}{11}\times 4\\-\frac{10}{11}\left(-5\right)+\frac{15}{11}\times 4\end{matrix}\right)$$

Mahia ngā tātaitanga.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}6\\10\end{matrix}\right)$$

Tangohia ngā huānga poukapa x me y.

$$x=6,y=10$$

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.

$$0.5x-0.8y+9=4,\frac{1}{3}x+\frac{1}{5}y=4$$

Kia ōrite ai a \frac{x}{2} me \frac{x}{3}, whakareatia ngā kīanga tau katoa kei ia taha o te whārite tuatahi ki te \frac{1}{3} me ngā kīanga tau katoa kei ia taha o te whārite tuarua ki te 0.5.

$$\frac{1}{3}\times 0.5x+\frac{1}{3}\left(-0.8\right)y+\frac{1}{3}\times 9=\frac{1}{3}\times 4,0.5\times \frac{1}{3}x+0.5\times \frac{1}{5}y=0.5\times 4$$

Whakarūnātia.

$$\frac{1}{6}x-\frac{4}{15}y+3=\frac{4}{3},\frac{1}{6}x+\frac{1}{10}y=2$$

Me tango \frac{1}{6}x+\frac{1}{10}y=2 mai i \frac{1}{6}x-\frac{4}{15}y+3=\frac{4}{3} mā te tango i ngā kīanga tau ōrite i ia taha o te tohu ōrite.

$$\frac{1}{6}x-\frac{1}{6}x-\frac{4}{15}y-\frac{1}{10}y+3=\frac{4}{3}-2$$

Tāpiri \frac{x}{6} ki te -\frac{x}{6}. Ka whakakore atu ngā kupu \frac{x}{6} me -\frac{x}{6}, ka toe he whārite me tētahi taurangi kotahi ka taea te whakaoti.

$$-\frac{4}{15}y-\frac{1}{10}y+3=\frac{4}{3}-2$$

Tāpiri -\frac{4y}{15} ki te -\frac{y}{10}.

$$-\frac{11}{30}y+3=\frac{4}{3}-2$$

Tāpiri \frac{4}{3} ki te -2.

$$-\frac{11}{30}y+3=-\frac{2}{3}$$

Me tango 3 mai i ngā taha e rua o te whārite.

$$-\frac{11}{30}y=-\frac{11}{3}$$

Whakawehea ngā taha e rua o te whārite ki te -\frac{11}{30}, he ōrite ki te whakarea i ngā taha e rua ki te tau huripoki o te hautanga.

$$y=10$$

Whakaurua te 10 mō y ki \frac{1}{3}x+\frac{1}{5}y=4. I te mea kotahi anake te taurangi kei te whārite i puta, ka taea e koe te whakaoti mō x hāngai tonu.

$$\frac{1}{3}x+\frac{1}{5}\times 10=4$$

Whakareatia \frac{1}{5} ki te 10.

$$\frac{1}{3}x+2=4$$

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

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

Me whakarea ngā taha e rua ki te 3.

$$x=6$$

Kua oti te pūnaha te whakatau.

$$x=6,y=10$$