Réitigh {l}{ax+by=c}{a^2x+b^2y=c}

$$\left. \begin{array} { l } { a x + b y = c } \\ { a ^ { 2 } x + b ^ { 2 } y = c } \end{array} \right.$$

Réitigh do x,y. (complex solution)

$$\left\{\begin{matrix}x=-\frac{c\left(1-b\right)}{a\left(b-a\right)}\text{, }y=-\frac{c\left(a-1\right)}{b\left(b-a\right)}\text{, }&b\neq 0\text{ and }a\neq b\text{ and }a\neq 0\\x=-\frac{by-c}{a}\text{, }y\in \mathrm{C}\text{, }&\left(c=0\text{ and }b=0\text{ and }a\neq 0\right)\text{ or }\left(c=0\text{ and }a=b\text{ and }b\neq 0\right)\text{ or }\left(a=1\text{ and }b=1\right)\text{ or }\left(a=1\text{ and }b=0\right)\\x\in \mathrm{C}\text{, }y=0\text{, }&c=0\text{ and }a=0\\x\in \mathrm{C}\text{, }y=c\text{, }&b=1\text{ and }a=0\\x\in \mathrm{C}\text{, }y\in \mathrm{C}\text{, }&c=0\text{ and }b=0\text{ and }a=0\end{matrix}\right.$$

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.

$$ax+by=c,a^{2}x+b^{2}y=c$$

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

$$ax+by=c$$

Bain by ón dá thaobh den chothromóid.

$$ax=\left(-b\right)y+c$$

Roinn an dá thaobh faoi a.

$$x=\frac{1}{a}\left(\left(-b\right)y+c\right)$$

Méadaigh \frac{1}{a} faoi -by+c.

$$x=\left(-\frac{b}{a}\right)y+\frac{c}{a}$$

Cuir x in aonad \frac{-by+c}{a} sa chothromóid eile, a^{2}x+b^{2}y=c.

$$a^{2}\left(\left(-\frac{b}{a}\right)y+\frac{c}{a}\right)+b^{2}y=c$$

Méadaigh a^{2} faoi \frac{-by+c}{a}.

$$\left(-ab\right)y+ac+b^{2}y=c$$

Suimigh -bay le b^{2}y?

$$b\left(b-a\right)y+ac=c$$

Bain ca ón dá thaobh den chothromóid.

$$b\left(b-a\right)y=c-ac$$

Roinn an dá thaobh faoi b\left(b-a\right).

$$y=\frac{c\left(1-a\right)}{b\left(b-a\right)}$$

Cuir y in aonad \frac{c\left(1-a\right)}{b\left(b-a\right)} in x=\left(-\frac{b}{a}\right)y+\frac{c}{a}. Toisc nach bhfuil ach athróg amháin sa chothromóid a bheidh mar thoradh air, is féidir leat réiteach díreach a fháil do x.

$$x=\left(-\frac{b}{a}\right)\times \frac{c\left(1-a\right)}{b\left(b-a\right)}+\frac{c}{a}$$

Méadaigh -\frac{b}{a} faoi \frac{c\left(1-a\right)}{b\left(b-a\right)}.

$$x=-\frac{c\left(1-a\right)}{a\left(b-a\right)}+\frac{c}{a}$$

Suimigh \frac{c}{a} le -\frac{\left(1-a\right)c}{\left(b-a\right)a}?

$$x=\frac{c\left(b-1\right)}{a\left(b-a\right)}$$

Tá an córas réitithe anois.

$$x=\frac{c\left(b-1\right)}{a\left(b-a\right)},y=\frac{c\left(1-a\right)}{b\left(b-a\right)}$$

Céimeanna ag Baint Úsáid as Maitrísí

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

$$ax+by=c,a^{2}x+b^{2}y=c$$

Scríobh na cothromóidí i bhfoirm mhaitríse.

$$\left(\begin{matrix}a&b\\a^{2}&b^{2}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}c\\c\end{matrix}\right)$$

Iolraigh faoi chlé an chothromóid faoi mhaitrís inbhéartach \left(\begin{matrix}a&b\\a^{2}&b^{2}\end{matrix}\right).

$$inverse(\left(\begin{matrix}a&b\\a^{2}&b^{2}\end{matrix}\right))\left(\begin{matrix}a&b\\a^{2}&b^{2}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}a&b\\a^{2}&b^{2}\end{matrix}\right))\left(\begin{matrix}c\\c\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}a&b\\a^{2}&b^{2}\end{matrix}\right))\left(\begin{matrix}c\\c\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}a&b\\a^{2}&b^{2}\end{matrix}\right))\left(\begin{matrix}c\\c\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{b^{2}}{ab^{2}-ba^{2}}&-\frac{b}{ab^{2}-ba^{2}}\\-\frac{a^{2}}{ab^{2}-ba^{2}}&\frac{a}{ab^{2}-ba^{2}}\end{matrix}\right)\left(\begin{matrix}c\\c\end{matrix}\right)$$

Déan an uimhríocht.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{b}{a\left(b-a\right)}&-\frac{1}{a\left(b-a\right)}\\-\frac{a}{b\left(b-a\right)}&\frac{1}{b\left(b-a\right)}\end{matrix}\right)\left(\begin{matrix}c\\c\end{matrix}\right)$$

Méadaigh na maitrísí.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{b}{a\left(b-a\right)}c+\left(-\frac{1}{a\left(b-a\right)}\right)c\\\left(-\frac{a}{b\left(b-a\right)}\right)c+\frac{1}{b\left(b-a\right)}c\end{matrix}\right)$$

Déan an uimhríocht.

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

Asbhain na heilimintí maitríse x agus y.

$$x=\frac{c\left(b-1\right)}{a\left(b-a\right)},y=\frac{c\left(1-a\right)}{b\left(b-a\right)}$$

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.

$$ax+by=c,a^{2}x+b^{2}y=c$$

Chun ax agus a^{2}x a dhéanamh cothrom, méadaigh gach téarma ar gach taobh den chéad chothromóid faoi a^{2} agus gach téarma ar gach taobh den dara cothromóid faoi a.

$$a^{2}ax+a^{2}by=a^{2}c,aa^{2}x+ab^{2}y=ac$$

Simpligh.

$$a^{3}x+ba^{2}y=ca^{2},a^{3}x+ab^{2}y=ac$$

Dealaigh a^{3}x+ab^{2}y=ac ó a^{3}x+ba^{2}y=ca^{2} trí théarmaí cosúla ar gach taobh den comhartha cothrom le a dhealú.

$$a^{3}x+\left(-a^{3}\right)x+ba^{2}y+\left(-ab^{2}\right)y=ca^{2}-ac$$

Suimigh a^{3}x le -a^{3}x? Cuirtear na téarmaí a^{3}x agus -a^{3}x ar ceal, agus níl fágtha ach cothromóid nach bhfuil inti ach athróg amháin is féidir a réiteach.

$$ba^{2}y+\left(-ab^{2}\right)y=ca^{2}-ac$$

Suimigh a^{2}by le -ab^{2}y?

$$ab\left(a-b\right)y=ca^{2}-ac$$

Suimigh a^{2}c le -ac?

$$ab\left(a-b\right)y=ac\left(a-1\right)$$

Roinn an dá thaobh faoi ab\left(a-b\right).

$$y=\frac{c\left(a-1\right)}{b\left(a-b\right)}$$

Cuir y in aonad \frac{\left(-1+a\right)c}{b\left(a-b\right)} in a^{2}x+b^{2}y=c. Toisc nach bhfuil ach athróg amháin sa chothromóid a bheidh mar thoradh air, is féidir leat réiteach díreach a fháil do x.

$$a^{2}x+b^{2}\times \frac{c\left(a-1\right)}{b\left(a-b\right)}=c$$

Méadaigh b^{2} faoi \frac{\left(-1+a\right)c}{b\left(a-b\right)}.

$$a^{2}x+\frac{bc\left(a-1\right)}{a-b}=c$$

Bain \frac{b\left(-1+a\right)c}{a-b} ón dá thaobh den chothromóid.

$$a^{2}x=\frac{ac\left(1-b\right)}{a-b}$$

Roinn an dá thaobh faoi a^{2}.

$$x=\frac{c\left(1-b\right)}{a\left(a-b\right)}$$

Tá an córas réitithe anois.

$$x=\frac{c\left(1-b\right)}{a\left(a-b\right)},y=\frac{c\left(a-1\right)}{b\left(a-b\right)}$$

Réitigh do x,y.

$$\left\{\begin{matrix}x=-\frac{c\left(1-b\right)}{a\left(b-a\right)}\text{, }y=-\frac{c\left(a-1\right)}{b\left(b-a\right)}\text{, }&b\neq 0\text{ and }a\neq b\text{ and }a\neq 0\\x=-\frac{by-c}{a}\text{, }y\in \mathrm{R}\text{, }&\left(c=0\text{ and }b=0\text{ and }a\neq 0\right)\text{ or }\left(c=0\text{ and }a=b\text{ and }b\neq 0\right)\text{ or }\left(a=1\text{ and }b=1\right)\text{ or }\left(a=1\text{ and }b=0\right)\\x\in \mathrm{R}\text{, }y=0\text{, }&c=0\text{ and }a=0\text{ and }b\neq 1\text{ and }b\neq 0\\x\in \mathrm{R}\text{, }y=c\text{, }&b=1\text{ and }a=0\\x\in \mathrm{R}\text{, }y\in \mathrm{R}\text{, }&c=0\text{ and }b=0\text{ and }a=0\end{matrix}\right.$$

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.

$$ax+by=c,a^{2}x+b^{2}y=c$$

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

$$ax+by=c$$

Bain by ón dá thaobh den chothromóid.

$$ax=\left(-b\right)y+c$$

Roinn an dá thaobh faoi a.

$$x=\frac{1}{a}\left(\left(-b\right)y+c\right)$$

Méadaigh \frac{1}{a} faoi -by+c.

$$x=\left(-\frac{b}{a}\right)y+\frac{c}{a}$$

Cuir x in aonad \frac{-by+c}{a} sa chothromóid eile, a^{2}x+b^{2}y=c.

$$a^{2}\left(\left(-\frac{b}{a}\right)y+\frac{c}{a}\right)+b^{2}y=c$$

Méadaigh a^{2} faoi \frac{-by+c}{a}.

$$\left(-ab\right)y+ac+b^{2}y=c$$

Suimigh -bay le b^{2}y?

$$b\left(b-a\right)y+ac=c$$

Bain ca ón dá thaobh den chothromóid.

$$b\left(b-a\right)y=c-ac$$

Roinn an dá thaobh faoi b\left(-a+b\right).

$$y=\frac{c\left(1-a\right)}{b\left(b-a\right)}$$

Cuir y in aonad \frac{c\left(1-a\right)}{b\left(-a+b\right)} in x=\left(-\frac{b}{a}\right)y+\frac{c}{a}. Toisc nach bhfuil ach athróg amháin sa chothromóid a bheidh mar thoradh air, is féidir leat réiteach díreach a fháil do x.

$$x=\left(-\frac{b}{a}\right)\times \frac{c\left(1-a\right)}{b\left(b-a\right)}+\frac{c}{a}$$

Méadaigh -\frac{b}{a} faoi \frac{c\left(1-a\right)}{b\left(-a+b\right)}.

$$x=-\frac{c\left(1-a\right)}{a\left(b-a\right)}+\frac{c}{a}$$

Suimigh \frac{c}{a} le -\frac{\left(1-a\right)c}{\left(-a+b\right)a}?

$$x=\frac{c\left(b-1\right)}{a\left(b-a\right)}$$

Tá an córas réitithe anois.

$$x=\frac{c\left(b-1\right)}{a\left(b-a\right)},y=\frac{c\left(1-a\right)}{b\left(b-a\right)}$$

Céimeanna ag Baint Úsáid as Maitrísí

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

$$ax+by=c,a^{2}x+b^{2}y=c$$

Scríobh na cothromóidí i bhfoirm mhaitríse.

$$\left(\begin{matrix}a&b\\a^{2}&b^{2}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}c\\c\end{matrix}\right)$$

Iolraigh faoi chlé an chothromóid faoi mhaitrís inbhéartach \left(\begin{matrix}a&b\\a^{2}&b^{2}\end{matrix}\right).

$$inverse(\left(\begin{matrix}a&b\\a^{2}&b^{2}\end{matrix}\right))\left(\begin{matrix}a&b\\a^{2}&b^{2}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}a&b\\a^{2}&b^{2}\end{matrix}\right))\left(\begin{matrix}c\\c\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}a&b\\a^{2}&b^{2}\end{matrix}\right))\left(\begin{matrix}c\\c\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}a&b\\a^{2}&b^{2}\end{matrix}\right))\left(\begin{matrix}c\\c\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{b^{2}}{ab^{2}-ba^{2}}&-\frac{b}{ab^{2}-ba^{2}}\\-\frac{a^{2}}{ab^{2}-ba^{2}}&\frac{a}{ab^{2}-ba^{2}}\end{matrix}\right)\left(\begin{matrix}c\\c\end{matrix}\right)$$

Déan an uimhríocht.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{b}{a\left(b-a\right)}&-\frac{1}{a\left(b-a\right)}\\-\frac{a}{b\left(b-a\right)}&\frac{1}{b\left(b-a\right)}\end{matrix}\right)\left(\begin{matrix}c\\c\end{matrix}\right)$$

Méadaigh na maitrísí.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{b}{a\left(b-a\right)}c+\left(-\frac{1}{a\left(b-a\right)}\right)c\\\left(-\frac{a}{b\left(b-a\right)}\right)c+\frac{1}{b\left(b-a\right)}c\end{matrix}\right)$$

Déan an uimhríocht.

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

Asbhain na heilimintí maitríse x agus y.

$$x=\frac{c\left(b-1\right)}{a\left(b-a\right)},y=\frac{c\left(1-a\right)}{b\left(b-a\right)}$$

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.

$$ax+by=c,a^{2}x+b^{2}y=c$$

Chun ax agus a^{2}x a dhéanamh cothrom, méadaigh gach téarma ar gach taobh den chéad chothromóid faoi a^{2} agus gach téarma ar gach taobh den dara cothromóid faoi a.

$$a^{2}ax+a^{2}by=a^{2}c,aa^{2}x+ab^{2}y=ac$$

Simpligh.

$$a^{3}x+ba^{2}y=ca^{2},a^{3}x+ab^{2}y=ac$$

Dealaigh a^{3}x+ab^{2}y=ac ó a^{3}x+ba^{2}y=ca^{2} trí théarmaí cosúla ar gach taobh den comhartha cothrom le a dhealú.

$$a^{3}x+\left(-a^{3}\right)x+ba^{2}y+\left(-ab^{2}\right)y=ca^{2}-ac$$

Suimigh a^{3}x le -a^{3}x? Cuirtear na téarmaí a^{3}x agus -a^{3}x ar ceal, agus níl fágtha ach cothromóid nach bhfuil inti ach athróg amháin is féidir a réiteach.

$$ba^{2}y+\left(-ab^{2}\right)y=ca^{2}-ac$$

Suimigh a^{2}by le -ab^{2}y?

$$ab\left(a-b\right)y=ca^{2}-ac$$

Suimigh a^{2}c le -ac?

$$ab\left(a-b\right)y=ac\left(a-1\right)$$

Roinn an dá thaobh faoi ab\left(a-b\right).

$$y=\frac{c\left(a-1\right)}{b\left(a-b\right)}$$

Cuir y in aonad \frac{\left(-1+a\right)c}{b\left(a-b\right)} in a^{2}x+b^{2}y=c. Toisc nach bhfuil ach athróg amháin sa chothromóid a bheidh mar thoradh air, is féidir leat réiteach díreach a fháil do x.

$$a^{2}x+b^{2}\times \frac{c\left(a-1\right)}{b\left(a-b\right)}=c$$

Méadaigh b^{2} faoi \frac{\left(-1+a\right)c}{b\left(a-b\right)}.

$$a^{2}x+\frac{bc\left(a-1\right)}{a-b}=c$$

Bain \frac{b\left(-1+a\right)c}{a-b} ón dá thaobh den chothromóid.

$$a^{2}x=\frac{ac\left(1-b\right)}{a-b}$$

Roinn an dá thaobh faoi a^{2}.

$$x=\frac{c\left(1-b\right)}{a\left(a-b\right)}$$

Tá an córas réitithe anois.

$$x=\frac{c\left(1-b\right)}{a\left(a-b\right)},y=\frac{c\left(a-1\right)}{b\left(a-b\right)}$$