Lời giải:
\(\left\{\begin{matrix} 3x-y+1\vdots 7\\ 2x+3y-1\vdots 7\end{matrix}\right.\Rightarrow \left\{\begin{matrix} 3(3x-y+1)\vdots 7\\ 2x+3y-1\vdots 7\end{matrix}\right.\)
\(\Rightarrow 3(3x-y+1)+(2x+3y-1)\vdots 7\)
\(\Rightarrow 11x+2\vdots 7\)
\(\Rightarrow 11(x-3)+35\vdots 7\Rightarrow 11(x-3)\vdots 7\Rightarrow x-3\vdots 7\)
\(\Rightarrow x\) chia 7 dư $3$
Đặt $x=7k+3$ thì:
\(3x-y+1\vdots 7\)
\(\Rightarrow 3(7k+3)-y+1\vdots 7\)
\(\Rightarrow 21k+7+3-y\vdots 7\Rightarrow 3-y\vdots 7\)
\(\Rightarrow y-3\vdots 7\) hay $y$ chia $7$ dư $3$
Vậy $x,y$ chia $7$ đều dư $3$
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