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Question Number 62439 by mathsolverby Abdo last updated on 21/Jun/19

sove inside Z/3Z the systeme { ((5x+7y =10)),((2x+5y =8)) :}

$${sove}\:{inside}\:{Z}/\mathrm{3}{Z}\:{the}\:{systeme} \\ $$$$\begin{cases}{\mathrm{5}{x}+\mathrm{7}{y}\:=\mathrm{10}}\\{\mathrm{2}{x}+\mathrm{5}{y}\:=\mathrm{8}}\end{cases} \\ $$$$ \\ $$

Commented by arcana last updated on 22/Jun/19

Z/3Z=Z_3 ={0^� ,1^� ,2^� } 0^� ={...,−9,−6,−3,0,3,6,9,...} 1^� ={...,−10,−7,−4,1,4,7,10,...} 2^� ={...,−11,−8,−5,2,5,8,11,...} 2^� x+1^� y=1^� 2^� x+2^� y=2^� ⇒2^� y−y=2^� −1^� =1^� y=1^� ⇒2^� x+(1^� )=1^� 2^� x=0^� (cancelativ)⇒x=0^� cause (Z_3 ,+^� ,∙^� ) is a field

$$\mathbb{Z}/\mathrm{3}\mathbb{Z}=\mathbb{Z}_{\mathrm{3}} =\left\{\bar {\mathrm{0}},\bar {\mathrm{1}},\bar {\mathrm{2}}\right\} \\ $$$$\bar {\mathrm{0}}=\left\{...,−\mathrm{9},−\mathrm{6},−\mathrm{3},\mathrm{0},\mathrm{3},\mathrm{6},\mathrm{9},...\right\} \\ $$$$\bar {\mathrm{1}}=\left\{...,−\mathrm{10},−\mathrm{7},−\mathrm{4},\mathrm{1},\mathrm{4},\mathrm{7},\mathrm{10},...\right\} \\ $$$$\bar {\mathrm{2}}=\left\{...,−\mathrm{11},−\mathrm{8},−\mathrm{5},\mathrm{2},\mathrm{5},\mathrm{8},\mathrm{11},...\right\} \\ $$$$ \\ $$$$\bar {\mathrm{2}}{x}+\bar {\mathrm{1}}{y}=\bar {\mathrm{1}} \\ $$$$\bar {\mathrm{2}}{x}+\bar {\mathrm{2}}{y}=\bar {\mathrm{2}} \\ $$$$ \\ $$$$\Rightarrow\bar {\mathrm{2}}{y}−{y}=\bar {\mathrm{2}}−\bar {\mathrm{1}}=\bar {\mathrm{1}} \\ $$$${y}=\bar {\mathrm{1}}\Rightarrow\bar {\mathrm{2}}{x}+\left(\bar {\mathrm{1}}\right)=\bar {\mathrm{1}} \\ $$$$\bar {\mathrm{2}}{x}=\bar {\mathrm{0}}\:\left(\mathrm{cancelativ}\right)\Rightarrow{x}=\bar {\mathrm{0}}\:\mathrm{cause}\:\left(\mathbb{Z}_{\mathrm{3}} ,\bar {+},\bar {\centerdot}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{field} \\ $$$$ \\ $$$$ \\ $$

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