Solving a system of linear congruences with two variables Solve the system of congruences $\begin{cases}2x+7y \equiv 2 \pmod 5 \\ 3x-4y \equiv 11 \pmod {13} \end{cases}$
This is more complicated to solve than an ordinary system of congruences, since we have $x$ and $y$. Any ideas on how to find general solutions would help.

$$2x+7y\equiv 2\pmod{5}\iff 2x+2y\equiv 2\pmod{5}$$
$$\stackrel{:2}\iff x+y\equiv 1\pmod{5}\iff x\equiv 1-y\pmod{5}$$
$$\iff x=5k+(1-y)$$
$$3x-4y\equiv 11\pmod{13}\iff 3x\equiv 11+4y\equiv 24-9y\pmod{13}$$
$$\stackrel{:3}\iff x\equiv 8-3y\pmod{13}\iff 5k+(1-y)\equiv 8-3y\pmod{13}$$
$$\iff 5k\equiv 7-2y\equiv 20-15y\pmod{13}\stackrel{:5}\iff k\equiv 4-3y\pmod{13}$$
$$\iff k=13m+(4-3y)$$
$$x=5(13m+(4-3y))+(1-y)=65m+21-16y$$
All the solutions to your system of linear congruences are given by $$(x,y)=(65m+21-16y,y)$$ with $y,m\in\Bbb Z$.
