Whakaoti i te {l}{(A+B)1/2-B=3/4}{(2A+B)1/4-B=5/4}

$$\left\{ \begin{array} { l } { ( A + B ) \frac { 1 } { 2 } - B = \frac { 3 } { 4 } } \\ { ( 2 A + B ) \frac { 1 } { 4 } - B = \frac { 5 } { 4 } } \end{array} \right.$$

Whakaoti mō A, B

$$A=-\frac{1}{2}=-0.5$$

$$B=-2$$

Ngā Upane e Whakamahi ana i te Whakakapinga

Whakaarohia te whārite tuatahi. Whakamahia te āhuatanga tohatoha hei whakarea te A+B ki te \frac{1}{2}.

$$\frac{1}{2}A+\frac{1}{2}B-B=\frac{3}{4}$$

Pahekotia te \frac{1}{2}B me -B, ka -\frac{1}{2}B.

$$\frac{1}{2}A-\frac{1}{2}B=\frac{3}{4}$$

Whakaarohia te whārite tuarua. Whakamahia te āhuatanga tohatoha hei whakarea te 2A+B ki te \frac{1}{4}.

$$\frac{1}{2}A+\frac{1}{4}B-B=\frac{5}{4}$$

Pahekotia te \frac{1}{4}B me -B, ka -\frac{3}{4}B.

$$\frac{1}{2}A-\frac{3}{4}B=\frac{5}{4}$$

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.

$$\frac{1}{2}A-\frac{1}{2}B=\frac{3}{4},\frac{1}{2}A-\frac{3}{4}B=\frac{5}{4}$$

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

$$\frac{1}{2}A-\frac{1}{2}B=\frac{3}{4}$$

Me tāpiri \frac{B}{2} ki ngā taha e rua o te whārite.

$$\frac{1}{2}A=\frac{1}{2}B+\frac{3}{4}$$

Me whakarea ngā taha e rua ki te 2.

$$A=2\left(\frac{1}{2}B+\frac{3}{4}\right)$$

Whakareatia 2 ki te \frac{B}{2}+\frac{3}{4}.

$$A=B+\frac{3}{2}$$

Whakakapia te B+\frac{3}{2} mō te A ki tērā atu whārite, \frac{1}{2}A-\frac{3}{4}B=\frac{5}{4}.

$$\frac{1}{2}\left(B+\frac{3}{2}\right)-\frac{3}{4}B=\frac{5}{4}$$

Whakareatia \frac{1}{2} ki te B+\frac{3}{2}.

$$\frac{1}{2}B+\frac{3}{4}-\frac{3}{4}B=\frac{5}{4}$$

Tāpiri \frac{B}{2} ki te -\frac{3B}{4}.

$$-\frac{1}{4}B+\frac{3}{4}=\frac{5}{4}$$

Me tango \frac{3}{4} mai i ngā taha e rua o te whārite.

$$-\frac{1}{4}B=\frac{1}{2}$$

Me whakarea ngā taha e rua ki te -4.

$$B=-2$$

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

$$A=-2+\frac{3}{2}$$

Tāpiri \frac{3}{2} ki te -2.

$$A=-\frac{1}{2}$$

Kua oti te pūnaha te whakatau.

$$A=-\frac{1}{2},B=-2$$

Ngā Upane e Whakamahi ana i Ngā Poukapa

Whakaarohia te whārite tuatahi. Whakamahia te āhuatanga tohatoha hei whakarea te A+B ki te \frac{1}{2}.

$$\frac{1}{2}A+\frac{1}{2}B-B=\frac{3}{4}$$

Pahekotia te \frac{1}{2}B me -B, ka -\frac{1}{2}B.

$$\frac{1}{2}A-\frac{1}{2}B=\frac{3}{4}$$

Whakaarohia te whārite tuarua. Whakamahia te āhuatanga tohatoha hei whakarea te 2A+B ki te \frac{1}{4}.

$$\frac{1}{2}A+\frac{1}{4}B-B=\frac{5}{4}$$

Pahekotia te \frac{1}{4}B me -B, ka -\frac{3}{4}B.

$$\frac{1}{2}A-\frac{3}{4}B=\frac{5}{4}$$

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

$$\frac{1}{2}A-\frac{1}{2}B=\frac{3}{4},\frac{1}{2}A-\frac{3}{4}B=\frac{5}{4}$$

Tuhia ngā whārite ki te tikanga tātai poukapa.

$$\left(\begin{matrix}\frac{1}{2}&-\frac{1}{2}\\\frac{1}{2}&-\frac{3}{4}\end{matrix}\right)\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}\frac{3}{4}\\\frac{5}{4}\end{matrix}\right)$$

Whakarea mauī i te whārite ki te poukapa kōaro o \left(\begin{matrix}\frac{1}{2}&-\frac{1}{2}\\\frac{1}{2}&-\frac{3}{4}\end{matrix}\right).

$$inverse(\left(\begin{matrix}\frac{1}{2}&-\frac{1}{2}\\\frac{1}{2}&-\frac{3}{4}\end{matrix}\right))\left(\begin{matrix}\frac{1}{2}&-\frac{1}{2}\\\frac{1}{2}&-\frac{3}{4}\end{matrix}\right)\left(\begin{matrix}A\\B\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{2}&-\frac{1}{2}\\\frac{1}{2}&-\frac{3}{4}\end{matrix}\right))\left(\begin{matrix}\frac{3}{4}\\\frac{5}{4}\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}A\\B\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{2}&-\frac{1}{2}\\\frac{1}{2}&-\frac{3}{4}\end{matrix}\right))\left(\begin{matrix}\frac{3}{4}\\\frac{5}{4}\end{matrix}\right)$$

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

$$\left(\begin{matrix}A\\B\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{2}&-\frac{1}{2}\\\frac{1}{2}&-\frac{3}{4}\end{matrix}\right))\left(\begin{matrix}\frac{3}{4}\\\frac{5}{4}\end{matrix}\right)$$

Mō te poukapa 2\times 2 \left(\begin{matrix}a&b\\c&d\end{matrix}\right), ko te poukapa kōaro ko \left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right), kia tuhia anō ai te whārite poukapa hei rapanga whakarea poukapa.

$$\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}-\frac{\frac{3}{4}}{\frac{1}{2}\left(-\frac{3}{4}\right)-\left(-\frac{1}{2}\times \frac{1}{2}\right)}&-\frac{-\frac{1}{2}}{\frac{1}{2}\left(-\frac{3}{4}\right)-\left(-\frac{1}{2}\times \frac{1}{2}\right)}\\-\frac{\frac{1}{2}}{\frac{1}{2}\left(-\frac{3}{4}\right)-\left(-\frac{1}{2}\times \frac{1}{2}\right)}&\frac{\frac{1}{2}}{\frac{1}{2}\left(-\frac{3}{4}\right)-\left(-\frac{1}{2}\times \frac{1}{2}\right)}\end{matrix}\right)\left(\begin{matrix}\frac{3}{4}\\\frac{5}{4}\end{matrix}\right)$$

Mahia ngā tātaitanga.

$$\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}6&-4\\4&-4\end{matrix}\right)\left(\begin{matrix}\frac{3}{4}\\\frac{5}{4}\end{matrix}\right)$$

Whakareatia ngā poukapa.

$$\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}6\times \frac{3}{4}-4\times \frac{5}{4}\\4\times \frac{3}{4}-4\times \frac{5}{4}\end{matrix}\right)$$

Mahia ngā tātaitanga.

$$\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{2}\\-2\end{matrix}\right)$$

Tangohia ngā huānga poukapa A me B.

$$A=-\frac{1}{2},B=-2$$

Ngā Upane e Whakamahi i te Tangohanga

Whakaarohia te whārite tuatahi. Whakamahia te āhuatanga tohatoha hei whakarea te A+B ki te \frac{1}{2}.

$$\frac{1}{2}A+\frac{1}{2}B-B=\frac{3}{4}$$

Pahekotia te \frac{1}{2}B me -B, ka -\frac{1}{2}B.

$$\frac{1}{2}A-\frac{1}{2}B=\frac{3}{4}$$

Whakaarohia te whārite tuarua. Whakamahia te āhuatanga tohatoha hei whakarea te 2A+B ki te \frac{1}{4}.

$$\frac{1}{2}A+\frac{1}{4}B-B=\frac{5}{4}$$

Pahekotia te \frac{1}{4}B me -B, ka -\frac{3}{4}B.

$$\frac{1}{2}A-\frac{3}{4}B=\frac{5}{4}$$

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.

$$\frac{1}{2}A-\frac{1}{2}B=\frac{3}{4},\frac{1}{2}A-\frac{3}{4}B=\frac{5}{4}$$

Me tango \frac{1}{2}A-\frac{3}{4}B=\frac{5}{4} mai i \frac{1}{2}A-\frac{1}{2}B=\frac{3}{4} mā te tango i ngā kīanga tau ōrite i ia taha o te tohu ōrite.

$$\frac{1}{2}A-\frac{1}{2}A-\frac{1}{2}B+\frac{3}{4}B=\frac{3-5}{4}$$

Tāpiri \frac{A}{2} ki te -\frac{A}{2}. Ka whakakore atu ngā kupu \frac{A}{2} me -\frac{A}{2}, ka toe he whārite me tētahi taurangi kotahi ka taea te whakaoti.

$$-\frac{1}{2}B+\frac{3}{4}B=\frac{3-5}{4}$$

Tāpiri -\frac{B}{2} ki te \frac{3B}{4}.

$$\frac{1}{4}B=\frac{3-5}{4}$$

Tāpiri \frac{3}{4} ki te -\frac{5}{4} 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.

$$\frac{1}{4}B=-\frac{1}{2}$$

Me whakarea ngā taha e rua ki te 4.

$$B=-2$$

Whakaurua te -2 mō B ki \frac{1}{2}A-\frac{3}{4}B=\frac{5}{4}. I te mea kotahi anake te taurangi kei te whārite i puta, ka taea e koe te whakaoti mō A hāngai tonu.

$$\frac{1}{2}A-\frac{3}{4}\left(-2\right)=\frac{5}{4}$$

Whakareatia -\frac{3}{4} ki te -2.

$$\frac{1}{2}A+\frac{3}{2}=\frac{5}{4}$$

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

$$\frac{1}{2}A=-\frac{1}{4}$$

Me whakarea ngā taha e rua ki te 2.

$$A=-\frac{1}{2}$$

Kua oti te pūnaha te whakatau.

$$A=-\frac{1}{2},B=-2$$