Réitigh {l}{I:2x+5y=14}{pi-2x+3y=4}

$$\left. \begin{array} { l } { I : 2 x + 5 y = 14 } \\ { \pi - 2 x + 3 y = 4 } \end{array} \right.$$

Réitigh do x,y.

$$x=\frac{2\left(5\pi +22\right)}{3I+20}$$

$$y=\frac{56+4I-\pi I}{3I+20}$$

$$I\neq -\frac{20}{3}$$

Céimeanna ag Baint Úsáid Ionadú

Chun péire cothromóidí a réiteach ag baint úsáid as ionadú, réitigh ceann de na cothromóidí ar dtús le ceann de na hathróga a fháil. Ansin ionadaigh an toradh don athróg sin sa chothromóid eile.

$$\frac{I}{2}x+5y=14,-2x+3y+\pi =4$$

Roghnaigh ceann de na cothromóidí agus réitigh é do x trí x ar an taobh clé den chomhartha ‘Cothrom le’ a aonrú.

$$\frac{I}{2}x+5y=14$$

Bain 5y ón dá thaobh den chothromóid.

$$\frac{I}{2}x=-5y+14$$

Roinn an dá thaobh faoi \frac{I}{2}.

$$x=\frac{2}{I}\left(-5y+14\right)$$

Méadaigh \frac{2}{I} faoi -5y+14.

$$x=\left(-\frac{10}{I}\right)y+\frac{28}{I}$$

Cuir x in aonad \frac{2\left(14-5y\right)}{I} sa chothromóid eile, -2x+3y+\pi =4.

$$-2\left(\left(-\frac{10}{I}\right)y+\frac{28}{I}\right)+3y+\pi =4$$

Méadaigh -2 faoi \frac{2\left(14-5y\right)}{I}.

$$\frac{20}{I}y-\frac{56}{I}+3y+\pi =4$$

Suimigh \frac{20y}{I} le 3y?

$$\left(3+\frac{20}{I}\right)y-\frac{56}{I}+\pi =4$$

Suimigh -\frac{56}{I} le \pi ?

$$\left(3+\frac{20}{I}\right)y+\pi -\frac{56}{I}=4$$

Bain \pi -\frac{56}{I} ón dá thaobh den chothromóid.

$$\left(3+\frac{20}{I}\right)y=4-\pi +\frac{56}{I}$$

Roinn an dá thaobh faoi 3+\frac{20}{I}.

$$y=\frac{56+4I-\pi I}{3I+20}$$

Cuir y in aonad \frac{56-I\pi +4I}{20+3I} in x=\left(-\frac{10}{I}\right)y+\frac{28}{I}. Toisc nach bhfuil ach athróg amháin sa chothromóid a bheidh mar thoradh air, is féidir leat réiteach díreach a fháil do x.

$$x=\left(-\frac{10}{I}\right)\times \frac{56+4I-\pi I}{3I+20}+\frac{28}{I}$$

Méadaigh -\frac{10}{I} faoi \frac{56-I\pi +4I}{20+3I}.

$$x=-\frac{10\left(56+4I-\pi I\right)}{I\left(3I+20\right)}+\frac{28}{I}$$

Suimigh \frac{28}{I} le -\frac{10\left(56-I\pi +4I\right)}{I\left(20+3I\right)}?

$$x=\frac{2\left(5\pi +22\right)}{3I+20}$$

Tá an córas réitithe anois.

$$x=\frac{2\left(5\pi +22\right)}{3I+20},y=\frac{56+4I-\pi I}{3I+20}$$

Céimeanna ag Baint Úsáid as Maitrísí

Cuir na cothromóidí i bhfoirm chaighdeánach agus ansin úsáid maitrísí chun córas na gcothromóidí a réiteach.

$$\frac{I}{2}x+5y=14,-2x+3y+\pi =4$$

Scríobh na cothromóidí i bhfoirm mhaitríse.

$$\left(\begin{matrix}\frac{I}{2}&5\\-2&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}14\\4-\pi \end{matrix}\right)$$

Iolraigh faoi chlé an chothromóid faoi mhaitrís inbhéartach \left(\begin{matrix}\frac{I}{2}&5\\-2&3\end{matrix}\right).

$$inverse(\left(\begin{matrix}\frac{I}{2}&5\\-2&3\end{matrix}\right))\left(\begin{matrix}\frac{I}{2}&5\\-2&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{I}{2}&5\\-2&3\end{matrix}\right))\left(\begin{matrix}14\\4-\pi \end{matrix}\right)$$

Is ionann an mhaitrís chéannachta agus toradh na maitríse agus a hinbhéarta.

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{I}{2}&5\\-2&3\end{matrix}\right))\left(\begin{matrix}14\\4-\pi \end{matrix}\right)$$

Iolraigh na maitrísí ar thaobh na láimhe clé den chomhartha ‘Cothrom le’.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{I}{2}&5\\-2&3\end{matrix}\right))\left(\begin{matrix}14\\4-\pi \end{matrix}\right)$$

Don mhaitrís 2\times 2 \left(\begin{matrix}a&b\\c&d\end{matrix}\right), is é an mhaitrís inbhéarta \left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right), mar sin is féidir cothromóid na maitríse a athscríobh mar fhadhb iolraithe maitríse.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{\frac{I}{2}\times 3-5\left(-2\right)}&-\frac{5}{\frac{I}{2}\times 3-5\left(-2\right)}\\-\frac{-2}{\frac{I}{2}\times 3-5\left(-2\right)}&\frac{I}{2\left(\frac{I}{2}\times 3-5\left(-2\right)\right)}\end{matrix}\right)\left(\begin{matrix}14\\4-\pi \end{matrix}\right)$$

Déan an uimhríocht.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{6}{3I+20}&-\frac{10}{3I+20}\\\frac{4}{3I+20}&\frac{I}{3I+20}\end{matrix}\right)\left(\begin{matrix}14\\4-\pi \end{matrix}\right)$$

Méadaigh na maitrísí.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{6}{3I+20}\times 14+\left(-\frac{10}{3I+20}\right)\left(4-\pi \right)\\\frac{4}{3I+20}\times 14+\frac{I}{3I+20}\left(4-\pi \right)\end{matrix}\right)$$

Déan an uimhríocht.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2\left(5\pi +22\right)}{3I+20}\\\frac{56+4I-\pi I}{3I+20}\end{matrix}\right)$$

Asbhain na heilimintí maitríse x agus y.

$$x=\frac{2\left(5\pi +22\right)}{3I+20},y=\frac{56+4I-\pi I}{3I+20}$$

Céimeanna ag Baint Úsáid as Díbirt

Chun réiteach a fháil trí dhíbirt, ní mór do chomhéifeachtaí ceann de na hathróga a bheith mar an gcéanna sa dá chothromóid ionas go gcealófar an athróg nuair a bhaintear cothromóid amháin ón gceann eile.

$$\frac{I}{2}x+5y=14,-2x+3y+\pi =4$$

Chun \frac{Ix}{2} agus -2x a dhéanamh cothrom, méadaigh gach téarma ar gach taobh den chéad chothromóid faoi -2 agus gach téarma ar gach taobh den dara cothromóid faoi \frac{1}{2}I.

$$-2\times \frac{I}{2}x-2\times 5y=-2\times 14,\frac{I}{2}\left(-2\right)x+\frac{I}{2}\times 3y+\frac{I}{2}\pi =\frac{I}{2}\times 4$$

Simpligh.

$$\left(-I\right)x-10y=-28,\left(-I\right)x+\frac{3I}{2}y+\frac{\pi I}{2}=2I$$

Dealaigh \left(-I\right)x+\frac{3I}{2}y+\frac{\pi I}{2}=2I ó \left(-I\right)x-10y=-28 trí théarmaí cosúla ar gach taobh den comhartha cothrom le a dhealú.

$$\left(-I\right)x+Ix-10y+\left(-\frac{3I}{2}\right)y-\frac{\pi I}{2}=-28-2I$$

Suimigh -Ix le Ix? Cuirtear na téarmaí -Ix agus Ix ar ceal, agus níl fágtha ach cothromóid nach bhfuil inti ach athróg amháin is féidir a réiteach.

$$-10y+\left(-\frac{3I}{2}\right)y-\frac{\pi I}{2}=-28-2I$$

Suimigh -10y le -\frac{3Iy}{2}?

$$\left(-\frac{3I}{2}-10\right)y-\frac{\pi I}{2}=-28-2I$$

Suimigh -28 le -2I?

$$\left(-\frac{3I}{2}-10\right)y-\frac{\pi I}{2}=-2I-28$$

Cuir \frac{I\pi }{2} leis an dá thaobh den chothromóid.

$$\left(-\frac{3I}{2}-10\right)y=\frac{\pi I}{2}-2I-28$$

Roinn an dá thaobh faoi -10-\frac{3I}{2}.

$$y=-\frac{\pi I-4I-56}{3I+20}$$

Cuir y in aonad -\frac{-56-4I+I\pi }{20+3I} in -2x+3y+\pi =4. Toisc nach bhfuil ach athróg amháin sa chothromóid a bheidh mar thoradh air, is féidir leat réiteach díreach a fháil do x.

$$-2x+3\left(-\frac{\pi I-4I-56}{3I+20}\right)+\pi =4$$

Méadaigh 3 faoi -\frac{-56-4I+I\pi }{20+3I}.

$$-2x-\frac{3\left(\pi I-4I-56\right)}{3I+20}+\pi =4$$

Suimigh -\frac{3\left(-56-4I+I\pi \right)}{20+3I} le \pi ?

$$-2x+\frac{4\left(3I+5\pi +42\right)}{3I+20}=4$$

Bain \frac{4\left(5\pi +3I+42\right)}{20+3I} ón dá thaobh den chothromóid.

$$-2x=-\frac{4\left(5\pi +22\right)}{3I+20}$$

Roinn an dá thaobh faoi -2.

$$x=\frac{2\left(5\pi +22\right)}{3I+20}$$

Tá an córas réitithe anois.

$$x=\frac{2\left(5\pi +22\right)}{3I+20},y=-\frac{\pi I-4I-56}{3I+20}$$