I'm new here , i want to solve the below system which $(x, y)$ are a paire in $\mathbb{N²}$.
$2x+2y= 1\bmod 10,4x+y= 7\bmod 10$
Thank you for your help
2026-04-02 12:24:47.1775132687
How do I solve the below system of congruence?
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The first congruence $2x+2y=1\pmod{10}$ doesn't have any solution. If $(a,b)$ is a solution, $1=10k-2a-2b$. RHS is even, LHS is odd.