Leysa bx+cy=a+b,ax[1/a-b-1/a+b]+cy[1/b-a-1/b+a]=frac{2a}{a+b}

$$b x + c y = a + b , a x [ \frac { 1 } { a - b } - \frac { 1 } { a + b } ] + c y [ \frac { 1 } { b - a } - \frac { 1 } { b + a } ] = \frac { 2 a } { a + b }$$

Leystu fyrir x, y (complex solution)

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

Skref með innsetningu

Til að leysa jöfnupar með innsetningu skal fyrst leysa eina jöfnuna fyrir eina breytuna. Síðan skal setja niðurstöðuna inn fyrir breytuna í hinni jöfnunni.

$$bx+cy=a+b,\left(-\frac{1}{a+b}+\frac{1}{a-b}\right)ax+\left(-\frac{1}{a+b}+\frac{1}{b-a}\right)cy=\frac{2a}{a+b}$$

Veldu eina jöfnuna og leystu x með því að einangra x vinstra megin við samasemmerkið.

$$bx+cy=a+b$$

Dragðu cy frá báðum hliðum jöfnunar.

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

Deildu báðum hliðum með b.

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

Margfaldaðu \frac{1}{b} sinnum -cy+a+b.

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

Settu \frac{-cy+a+b}{b} inn fyrir x í hinni jöfnunni, \left(-\frac{1}{a+b}+\frac{1}{a-b}\right)ax+\left(-\frac{1}{a+b}+\frac{1}{b-a}\right)cy=\frac{2a}{a+b}.

$$\left(-\frac{1}{a+b}+\frac{1}{a-b}\right)a\left(\left(-\frac{c}{b}\right)y+\frac{a+b}{b}\right)+\left(-\frac{1}{a+b}+\frac{1}{b-a}\right)cy=\frac{2a}{a+b}$$

Margfaldaðu a\left(\left(a-b\right)^{-1}-\left(a+b\right)^{-1}\right) sinnum \frac{-cy+a+b}{b}.

$$\left(-\frac{2ac}{\left(a-b\right)\left(a+b\right)}\right)y+\frac{2a}{a-b}+\left(-\frac{1}{a+b}+\frac{1}{b-a}\right)cy=\frac{2a}{a+b}$$

Leggðu -\frac{2acy}{\left(a-b\right)\left(a+b\right)} saman við \frac{2cay}{\left(b-a\right)\left(b+a\right)}.

$$\frac{4ac}{\left(b-a\right)\left(a+b\right)}y+\frac{2a}{a-b}=\frac{2a}{a+b}$$

Dragðu \frac{2a}{a-b} frá báðum hliðum jöfnunar.

$$\frac{4ac}{\left(b-a\right)\left(a+b\right)}y=-\frac{4ab}{a^{2}-b^{2}}$$

Deildu báðum hliðum með \frac{4ca}{\left(b-a\right)\left(a+b\right)}.

$$y=\frac{b}{c}$$

Skiptu \frac{b}{c} út fyrir y í x=\left(-\frac{c}{b}\right)y+\frac{a+b}{b}. Þar sem jafnan sem af þessu leiðir inniheldur einungis eina breytu geturðu leyst x strax.

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

Margfaldaðu -\frac{c}{b} sinnum \frac{b}{c}.

$$x=-1+\frac{a+b}{b}$$

Leggðu \frac{a+b}{b} saman við -1.

$$x=\frac{a}{b}$$

Leyst var úr kerfinu.

$$x=\frac{a}{b},y=\frac{b}{c}$$

Skref með fylkjum

Settu jöfnurnar í staðlað form og notaðu svo fylki til að leysa jöfnuhneppið.

$$bx+cy=a+b,\left(-\frac{1}{a+b}+\frac{1}{a-b}\right)ax+\left(-\frac{1}{a+b}+\frac{1}{b-a}\right)cy=\frac{2a}{a+b}$$

Skrifaðu jöfnurnar á fylkjaformi.

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

Margfaldaðu vinstri hlið jöfnunnar með andhverfu fylkis \left(\begin{matrix}b&c\\-\frac{2ab}{\left(-a+b\right)\left(a+b\right)}&\frac{2ca}{\left(b-a\right)\left(b+a\right)}\end{matrix}\right).

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

Margfeldi fylkis og andhverfu þess er einingarfylki.

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}b&c\\\frac{2ab}{\left(a-b\right)\left(a+b\right)}&\frac{2ac}{\left(b-a\right)\left(a+b\right)}\end{matrix}\right))\left(\begin{matrix}a+b\\\frac{2a}{a+b}\end{matrix}\right)$$

Margfaldaðu fylkin vinstra megin við samasemmerkið.

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

Fyrir 2\times 2-fylkið \left(\begin{matrix}a&b\\c&d\end{matrix}\right) er andhverfa fylkið \left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right), þannig að hægt er að endurrita fylkisjöfnuna sem fylkismargföldunardæmi.

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

Reiknaðu.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2b}&\frac{a}{4b}-\frac{b}{4a}\\\frac{1}{2c}&\frac{\left(b-a\right)\left(a+b\right)}{4ac}\end{matrix}\right)\left(\begin{matrix}a+b\\\frac{2a}{a+b}\end{matrix}\right)$$

Margfaldaðu fylkin.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2b}\left(a+b\right)+\left(\frac{a}{4b}-\frac{b}{4a}\right)\times \frac{2a}{a+b}\\\frac{1}{2c}\left(a+b\right)+\frac{\left(b-a\right)\left(a+b\right)}{4ac}\times \frac{2a}{a+b}\end{matrix}\right)$$

Reiknaðu.

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

Dragðu út stuðul fylkjanna x og y.

$$x=\frac{a}{b},y=\frac{b}{c}$$

Skref með útilokun

Til að nota útilokun við lausn verða stuðlar einnar breytunnar að vera eins í báðum jöfnunum til að breytan núllist út þegar ein jafna er dregin frá annarri.

$$bx+cy=a+b,\left(-\frac{1}{a+b}+\frac{1}{a-b}\right)ax+\left(-\frac{1}{a+b}+\frac{1}{b-a}\right)cy=\frac{2a}{a+b}$$

Til að gera bx og \frac{2abx}{\left(a-b\right)\left(a+b\right)} jafnt skal margfalda alla liði á hverri hlið fyrstu jöfnunnar með a\left(\left(a-b\right)^{-1}-\left(a+b\right)^{-1}\right) og alla liði á hverri hlið annarrar jöfnunnar með b.

$$\left(-\frac{1}{a+b}+\frac{1}{a-b}\right)abx+\left(-\frac{1}{a+b}+\frac{1}{a-b}\right)acy=\left(-\frac{1}{a+b}+\frac{1}{a-b}\right)a\left(a+b\right),b\left(-\frac{1}{a+b}+\frac{1}{a-b}\right)ax+b\left(-\frac{1}{a+b}+\frac{1}{b-a}\right)cy=b\times \frac{2a}{a+b}$$

Einfaldaðu.

$$\frac{2ab^{2}}{\left(a-b\right)\left(a+b\right)}x+\frac{2abc}{\left(a-b\right)\left(a+b\right)}y=\frac{2ab}{a-b},\frac{2ab^{2}}{\left(a-b\right)\left(a+b\right)}x+\frac{2abc}{\left(b-a\right)\left(a+b\right)}y=\frac{2ab}{a+b}$$

Dragðu \frac{2ab^{2}}{\left(a-b\right)\left(a+b\right)}x+\frac{2abc}{\left(b-a\right)\left(a+b\right)}y=\frac{2ab}{a+b} frá \frac{2ab^{2}}{\left(a-b\right)\left(a+b\right)}x+\frac{2abc}{\left(a-b\right)\left(a+b\right)}y=\frac{2ab}{a-b} með því að draga frá líka liði sitt hvorum megin við samasemmerkið.

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

Leggðu \frac{2ab^{2}x}{\left(a-b\right)\left(a+b\right)} saman við -\frac{2ab^{2}x}{\left(a-b\right)\left(a+b\right)}. Liðirnir \frac{2ab^{2}x}{\left(a-b\right)\left(a+b\right)} og -\frac{2ab^{2}x}{\left(a-b\right)\left(a+b\right)} núlla hvorn annan út, sem skilur eftir jöfnu með einungis eina breytu sem hægt er að leysa.

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

Leggðu \frac{2abcy}{\left(a-b\right)\left(a+b\right)} saman við -\frac{2bcay}{\left(b-a\right)\left(b+a\right)}.

$$\frac{4abc}{\left(a-b\right)\left(a+b\right)}y=\frac{2ab}{a-b}-\frac{2ab}{a+b}$$

Leggðu \frac{2ab}{a-b} saman við -\frac{2ba}{a+b}.

$$\frac{4abc}{\left(a-b\right)\left(a+b\right)}y=\frac{4ab^{2}}{\left(a-b\right)\left(a+b\right)}$$

Deildu báðum hliðum með \frac{4bca}{\left(a-b\right)\left(a+b\right)}.

$$y=\frac{b}{c}$$

Skiptu \frac{b}{c} út fyrir y í \left(-\frac{1}{a+b}+\frac{1}{a-b}\right)ax+\left(-\frac{1}{a+b}+\frac{1}{b-a}\right)cy=\frac{2a}{a+b}. Þar sem jafnan sem af þessu leiðir inniheldur einungis eina breytu geturðu leyst x strax.

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

Margfaldaðu c\left(\left(b-a\right)^{-1}-\left(b+a\right)^{-1}\right) sinnum \frac{b}{c}.

$$\left(-\frac{1}{a+b}+\frac{1}{a-b}\right)ax+\frac{2ab}{\left(b-a\right)\left(a+b\right)}=\frac{2a}{a+b}$$

Dragðu \frac{2ab}{\left(b-a\right)\left(b+a\right)} frá báðum hliðum jöfnunar.

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

Deildu báðum hliðum með a\left(\left(a-b\right)^{-1}-\left(a+b\right)^{-1}\right).

$$x=\frac{a}{b}$$

Leyst var úr kerfinu.

$$x=\frac{a}{b},y=\frac{b}{c}$$

Leystu fyrir x, y

$$\left\{\begin{matrix}x=\frac{a}{b}\text{, }y=\frac{b}{c}\text{, }&c\neq 0\text{ and }b\neq 0\text{ and }|b|\neq |a|\\x=\frac{b-cy}{b}\text{, }y\in \mathrm{R}\text{, }&a=0\text{ and }b\neq 0\end{matrix}\right.$$

Skref með innsetningu

Til að leysa jöfnupar með innsetningu skal fyrst leysa eina jöfnuna fyrir eina breytuna. Síðan skal setja niðurstöðuna inn fyrir breytuna í hinni jöfnunni.

$$bx+cy=a+b,\left(-\frac{1}{a+b}+\frac{1}{a-b}\right)ax+\left(-\frac{1}{a+b}+\frac{1}{b-a}\right)cy=\frac{2a}{a+b}$$

Veldu eina jöfnuna og leystu x með því að einangra x vinstra megin við samasemmerkið.

$$bx+cy=a+b$$

Dragðu cy frá báðum hliðum jöfnunar.

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

Deildu báðum hliðum með b.

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

Margfaldaðu \frac{1}{b} sinnum -cy+a+b.

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

Settu \frac{-cy+a+b}{b} inn fyrir x í hinni jöfnunni, \left(-\frac{1}{a+b}+\frac{1}{a-b}\right)ax+\left(-\frac{1}{a+b}+\frac{1}{b-a}\right)cy=\frac{2a}{a+b}.

$$\left(-\frac{1}{a+b}+\frac{1}{a-b}\right)a\left(\left(-\frac{c}{b}\right)y+\frac{a+b}{b}\right)+\left(-\frac{1}{a+b}+\frac{1}{b-a}\right)cy=\frac{2a}{a+b}$$

Margfaldaðu a\left(\left(a-b\right)^{-1}-\left(a+b\right)^{-1}\right) sinnum \frac{-cy+a+b}{b}.

$$\left(-\frac{2ac}{\left(a-b\right)\left(a+b\right)}\right)y+\frac{2a}{a-b}+\left(-\frac{1}{a+b}+\frac{1}{b-a}\right)cy=\frac{2a}{a+b}$$

Leggðu -\frac{2acy}{\left(a-b\right)\left(a+b\right)} saman við \frac{2cay}{\left(b-a\right)\left(b+a\right)}.

$$\frac{4ac}{\left(b-a\right)\left(a+b\right)}y+\frac{2a}{a-b}=\frac{2a}{a+b}$$

Dragðu \frac{2a}{a-b} frá báðum hliðum jöfnunar.

$$\frac{4ac}{\left(b-a\right)\left(a+b\right)}y=-\frac{4ab}{a^{2}-b^{2}}$$

Deildu báðum hliðum með \frac{4ca}{\left(b-a\right)\left(a+b\right)}.

$$y=\frac{b}{c}$$

Skiptu \frac{b}{c} út fyrir y í x=\left(-\frac{c}{b}\right)y+\frac{a+b}{b}. Þar sem jafnan sem af þessu leiðir inniheldur einungis eina breytu geturðu leyst x strax.

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

Margfaldaðu -\frac{c}{b} sinnum \frac{b}{c}.

$$x=-1+\frac{a+b}{b}$$

Leggðu \frac{a+b}{b} saman við -1\text{, }|b|\neq |a|.

$$x=\frac{a}{b}$$

Leyst var úr kerfinu.

$$x=\frac{a}{b},y=\frac{b}{c}$$

Skref með fylkjum

Settu jöfnurnar í staðlað form og notaðu svo fylki til að leysa jöfnuhneppið.

$$bx+cy=a+b,\left(-\frac{1}{a+b}+\frac{1}{a-b}\right)ax+\left(-\frac{1}{a+b}+\frac{1}{b-a}\right)cy=\frac{2a}{a+b}$$

Skrifaðu jöfnurnar á fylkjaformi.

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

Margfaldaðu vinstri hlið jöfnunnar með andhverfu fylkis \left(\begin{matrix}b&c\\-\frac{2ab}{\left(-a+b\right)\left(a+b\right)}&\frac{2ca}{\left(b-a\right)\left(b+a\right)}\end{matrix}\right).

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

Margfeldi fylkis og andhverfu þess er einingarfylki.

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}b&c\\\frac{2ab}{\left(a-b\right)\left(a+b\right)}&\frac{2ac}{\left(b-a\right)\left(a+b\right)}\end{matrix}\right))\left(\begin{matrix}a+b\\\frac{2a}{a+b}\end{matrix}\right)$$

Margfaldaðu fylkin vinstra megin við samasemmerkið.

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

Fyrir 2\times 2-fylkið \left(\begin{matrix}a&b\\c&d\end{matrix}\right) er andhverfa fylkið \left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right), þannig að hægt er að endurrita fylkisjöfnuna sem fylkismargföldunardæmi.

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

Reiknaðu.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2b}&\frac{a}{4b}-\frac{b}{4a}\\\frac{1}{2c}&\frac{\left(b-a\right)\left(a+b\right)}{4ac}\end{matrix}\right)\left(\begin{matrix}a+b\\\frac{2a}{a+b}\end{matrix}\right)$$

Margfaldaðu fylkin.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2b}\left(a+b\right)+\left(\frac{a}{4b}-\frac{b}{4a}\right)\times \frac{2a}{a+b}\\\frac{1}{2c}\left(a+b\right)+\frac{\left(b-a\right)\left(a+b\right)}{4ac}\times \frac{2a}{a+b}\end{matrix}\right)$$

Reiknaðu.

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

Dragðu út stuðul fylkjanna x og y.

$$x=\frac{a}{b},y=\frac{b}{c}$$

Skref með útilokun

Til að nota útilokun við lausn verða stuðlar einnar breytunnar að vera eins í báðum jöfnunum til að breytan núllist út þegar ein jafna er dregin frá annarri.

$$bx+cy=a+b,\left(-\frac{1}{a+b}+\frac{1}{a-b}\right)ax+\left(-\frac{1}{a+b}+\frac{1}{b-a}\right)cy=\frac{2a}{a+b}$$

Til að gera bx og \frac{2abx}{\left(a-b\right)\left(a+b\right)} jafnt skal margfalda alla liði á hverri hlið fyrstu jöfnunnar með a\left(\left(a-b\right)^{-1}-\left(a+b\right)^{-1}\right) og alla liði á hverri hlið annarrar jöfnunnar með b.

$$\left(-\frac{1}{a+b}+\frac{1}{a-b}\right)abx+\left(-\frac{1}{a+b}+\frac{1}{a-b}\right)acy=\left(-\frac{1}{a+b}+\frac{1}{a-b}\right)a\left(a+b\right),b\left(-\frac{1}{a+b}+\frac{1}{a-b}\right)ax+b\left(-\frac{1}{a+b}+\frac{1}{b-a}\right)cy=b\times \frac{2a}{a+b}$$

Einfaldaðu.

$$\frac{2ab^{2}}{\left(a-b\right)\left(a+b\right)}x+\frac{2abc}{\left(a-b\right)\left(a+b\right)}y=\frac{2ab}{a-b},\frac{2ab^{2}}{\left(a-b\right)\left(a+b\right)}x+\frac{2abc}{\left(b-a\right)\left(a+b\right)}y=\frac{2ab}{a+b}$$

Dragðu \frac{2ab^{2}}{\left(a-b\right)\left(a+b\right)}x+\frac{2abc}{\left(b-a\right)\left(a+b\right)}y=\frac{2ab}{a+b} frá \frac{2ab^{2}}{\left(a-b\right)\left(a+b\right)}x+\frac{2abc}{\left(a-b\right)\left(a+b\right)}y=\frac{2ab}{a-b} með því að draga frá líka liði sitt hvorum megin við samasemmerkið.

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

Leggðu \frac{2ab^{2}x}{\left(a-b\right)\left(a+b\right)} saman við -\frac{2ab^{2}x}{\left(a-b\right)\left(a+b\right)}. Liðirnir \frac{2ab^{2}x}{\left(a-b\right)\left(a+b\right)} og -\frac{2ab^{2}x}{\left(a-b\right)\left(a+b\right)} núlla hvorn annan út, sem skilur eftir jöfnu með einungis eina breytu sem hægt er að leysa.

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

Leggðu \frac{2abcy}{\left(a-b\right)\left(a+b\right)} saman við -\frac{2bcay}{\left(b-a\right)\left(b+a\right)}.

$$\frac{4abc}{\left(a-b\right)\left(a+b\right)}y=\frac{2ab}{a-b}-\frac{2ab}{a+b}$$

Leggðu \frac{2ab}{a-b} saman við -\frac{2ba}{a+b}.

$$\frac{4abc}{\left(a-b\right)\left(a+b\right)}y=\frac{4ab^{2}}{\left(a-b\right)\left(a+b\right)}$$

Deildu báðum hliðum með \frac{4bca}{\left(a-b\right)\left(a+b\right)}.

$$y=\frac{b}{c}$$

Skiptu \frac{b}{c} út fyrir y í \left(-\frac{1}{a+b}+\frac{1}{a-b}\right)ax+\left(-\frac{1}{a+b}+\frac{1}{b-a}\right)cy=\frac{2a}{a+b}. Þar sem jafnan sem af þessu leiðir inniheldur einungis eina breytu geturðu leyst x strax.

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

Margfaldaðu c\left(\left(b-a\right)^{-1}-\left(b+a\right)^{-1}\right) sinnum \frac{b}{c}.

$$\left(-\frac{1}{a+b}+\frac{1}{a-b}\right)ax+\frac{2ab}{\left(b-a\right)\left(a+b\right)}=\frac{2a}{a+b}$$

Dragðu \frac{2ab}{\left(b-a\right)\left(b+a\right)} frá báðum hliðum jöfnunar.

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

Deildu báðum hliðum með a\left(\left(a-b\right)^{-1}-\left(a+b\right)^{-1}\right).

$$x=\frac{a}{b}$$

Leyst var úr kerfinu.

$$x=\frac{a}{b},y=\frac{b}{c}$$