Past exam of the mathematics course of the University of Cambridge / 2015 / ia / Paper 4 / 1E / a

Subtracting twice the first linear congruence from the second gives $5x\equiv 1(\mathrm{mod}53)$. Since $5\cdot 32=3\cdot 53+1$, the modular inverse of $5$ modulo $53$ is $32$. Thus $x\equiv 32(\mathrm{mod}53)$. Substitution in the first linear congruence then gives $2y\equiv 23(\mathrm{mod}53)$, whence $y\equiv 38(\mathrm{mod}53)$. All integer solutions areThe two free integers are independent. Conversely, these values give $6x+2y\equiv 268\equiv 3$ and $17x+4y\equiv 696\equiv 7(\mathrm{mod}53)$, so no additional restriction is needed.