= Solution
Subtracting twice the first <linear congruence> from the second gives $5x\equiv1\pmod{53}$. Since $5\cdot32=3\cdot53+1$, the <modular inverse> of $5$ modulo $53$ is $32$. Thus $x\equiv32\pmod{53}$. Substitution in the first <linear congruence> then gives $2y\equiv23\pmod{53}$, whence $y\equiv38\pmod{53}$. \b[All integer solutions are]
$$
\boxed{x=32+53r,\qquad y=38+53t,\qquad r,t\in\mathbb Z.}
$$
The two free <integers> are independent. Conversely, these values give $6x+2y\equiv268\equiv3$ and $17x+4y\equiv696\equiv7\pmod{53}$, so no additional restriction is needed.
Back to article page
