Whakaoti i te {l}{2(2x-3)+3(y+4)=7}{4(x+2)-5(2-y)=-3}

$$\left\{ \begin{array} { l } { 2 ( 2 x - 3 ) + 3 ( y + 4 ) = 7 } \\ { 4 ( x + 2 ) - 5 ( 2 - y ) = - 3 } \end{array} \right.$$

Whakaoti mō x, y

$$x=1$$

$$y=-1$$

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.

$$2\left(2x-3\right)+3\left(y+4\right)=7,4\left(x+2\right)-5\left(-y+2\right)=-3$$

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

$$2\left(2x-3\right)+3\left(y+4\right)=7$$

Whakareatia 2 ki te 2x-3.

$$4x-6+3\left(y+4\right)=7$$

Whakareatia 3 ki te y+4.

$$4x-6+3y+12=7$$

Tāpiri -6 ki te 12.

$$4x+3y+6=7$$

Me tango 6 mai i ngā taha e rua o te whārite.

$$4x+3y=1$$

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

$$4x=-3y+1$$

Whakawehea ngā taha e rua ki te 4.

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

Whakareatia \frac{1}{4} ki te -3y+1.

$$x=-\frac{3}{4}y+\frac{1}{4}$$

Whakakapia te \frac{-3y+1}{4} mō te x ki tērā atu whārite, 4\left(x+2\right)-5\left(-y+2\right)=-3.

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

Tāpiri \frac{1}{4} ki te 2.

$$4\left(-\frac{3}{4}y+\frac{9}{4}\right)-5\left(-y+2\right)=-3$$

Whakareatia 4 ki te \frac{-3y+9}{4}.

$$-3y+9-5\left(-y+2\right)=-3$$

Whakareatia -5 ki te -y+2.

$$-3y+9+5y-10=-3$$

Tāpiri -3y ki te 5y.

$$2y+9-10=-3$$

Tāpiri 9 ki te -10.

$$2y-1=-3$$

Me tāpiri 1 ki ngā taha e rua o te whārite.

$$2y=-2$$

Whakawehea ngā taha e rua ki te 2.

$$y=-1$$

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

$$x=-\frac{3}{4}\left(-1\right)+\frac{1}{4}$$

Whakareatia -\frac{3}{4} ki te -1.

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

Tāpiri \frac{1}{4} ki te \frac{3}{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.

$$x=1$$

Kua oti te pūnaha te whakatau.

$$x=1,y=-1$$

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.

$$2\left(2x-3\right)+3\left(y+4\right)=7,4\left(x+2\right)-5\left(-y+2\right)=-3$$

Whakarūnātia te whārite tuatahi ki te āhua tānga ngahuru.

$$2\left(2x-3\right)+3\left(y+4\right)=7$$

Whakareatia 2 ki te 2x-3.

$$4x-6+3\left(y+4\right)=7$$

Whakareatia 3 ki te y+4.

$$4x-6+3y+12=7$$

Tāpiri -6 ki te 12.

$$4x+3y+6=7$$

Me tango 6 mai i ngā taha e rua o te whārite.

$$4x+3y=1$$

Whakarūnātia te whārite tuarua ki te āhua tānga ngahuru.

$$4\left(x+2\right)-5\left(-y+2\right)=-3$$

Whakareatia 4 ki te x+2.

$$4x+8-5\left(-y+2\right)=-3$$

Whakareatia -5 ki te -y+2.

$$4x+8+5y-10=-3$$

Tāpiri 8 ki te -10.

$$4x+5y-2=-3$$

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

$$4x+5y=-1$$

Tuhia ngā whārite ki te tikanga tātai poukapa.

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

Whakarea mauī i te whārite ki te poukapa kōaro o \left(\begin{matrix}4&3\\4&5\end{matrix}\right).

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

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

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

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

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

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.

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

Mahia ngā tātaitanga.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{8}&-\frac{3}{8}\\-\frac{1}{2}&\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}1\\-1\end{matrix}\right)$$

Whakareatia ngā poukapa.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{8}-\frac{3}{8}\left(-1\right)\\-\frac{1}{2}+\frac{1}{2}\left(-1\right)\end{matrix}\right)$$

Mahia ngā tātaitanga.

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

Tangohia ngā huānga poukapa x me y.

$$x=1,y=-1$$