<s>[INST] Find two solutions of $299x+247y=13$ Applying Euclidean algorithm I got 
$299=247+52$
$247=4\cdot 52+39$
$52=39+13$
$39=3\cdot 13+0$
Applying this in reverse order I got 
$52-39=13$
$52-(247-4\cdot 52)=13$
$5(299-247)-247=13$
$5\cdot 299-6\cdot 247=13$
$x=5, y=-6$, this is one solution, what is the other solution?
 [/INST] As $13|(299,247)=13$ 
from this, we shall have $13$ in-congruent solutions $\pmod{13}$
$299x+247y=13\iff23x+19y=1=23\cdot5-19\cdot6$
$\iff23(x-5)=-19(y+6)$
As $\dfrac{19(y+6)}{23}=5-x$ which is an integer,
$\implies23\mid(y+6)$ as $(19,23)=1$
$\iff y\equiv-6\pmod{23}$
and similarly, $x\equiv5\pmod{19}$
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