Question Number 62439 by mathsolverby Abdo last updated on 21/Jun/19
$${sove}\:{inside}\:{Z}/\mathrm{3}{Z}\:{the}\:{systeme} \\ $$$$\begin{cases}{\mathrm{5}{x}+\mathrm{7}{y}\:=\mathrm{10}}\\{\mathrm{2}{x}+\mathrm{5}{y}\:=\mathrm{8}}\end{cases} \\ $$$$ \\ $$
Commented by arcana last updated on 22/Jun/19
$$\mathbb{Z}/\mathrm{3}\mathbb{Z}=\mathbb{Z}_{\mathrm{3}} =\left\{\bar {\mathrm{0}},\bar {\mathrm{1}},\bar {\mathrm{2}}\right\} \\ $$$$\bar {\mathrm{0}}=\left\{…,−\mathrm{9},−\mathrm{6},−\mathrm{3},\mathrm{0},\mathrm{3},\mathrm{6},\mathrm{9},…\right\} \\ $$$$\bar {\mathrm{1}}=\left\{…,−\mathrm{10},−\mathrm{7},−\mathrm{4},\mathrm{1},\mathrm{4},\mathrm{7},\mathrm{10},…\right\} \\ $$$$\bar {\mathrm{2}}=\left\{…,−\mathrm{11},−\mathrm{8},−\mathrm{5},\mathrm{2},\mathrm{5},\mathrm{8},\mathrm{11},…\right\} \\ $$$$ \\ $$$$\bar {\mathrm{2}}{x}+\bar {\mathrm{1}}{y}=\bar {\mathrm{1}} \\ $$$$\bar {\mathrm{2}}{x}+\bar {\mathrm{2}}{y}=\bar {\mathrm{2}} \\ $$$$ \\ $$$$\Rightarrow\bar {\mathrm{2}}{y}−{y}=\bar {\mathrm{2}}−\bar {\mathrm{1}}=\bar {\mathrm{1}} \\ $$$${y}=\bar {\mathrm{1}}\Rightarrow\bar {\mathrm{2}}{x}+\left(\bar {\mathrm{1}}\right)=\bar {\mathrm{1}} \\ $$$$\bar {\mathrm{2}}{x}=\bar {\mathrm{0}}\:\left(\mathrm{cancelativ}\right)\Rightarrow{x}=\bar {\mathrm{0}}\:\mathrm{cause}\:\left(\mathbb{Z}_{\mathrm{3}} ,\bar {+},\bar {\centerdot}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{field} \\ $$$$ \\ $$$$ \\ $$