<s>[INST] Number Theory : Solve the system of congruence $28x+17y\equiv 18 \pmod{41}$ and $31x+11y\equiv 35\pmod{41}$ Number Theory : Solve the system of congruence
(1)    $28x+17y\equiv 18 \pmod{41}$
(2)   $31x+11y\equiv 35\pmod{41}$
Attempt :
we know that equation (1) and (2) are in the same$\pmod{41}$. so we can use Modular arithmetic
lets multiply the first equation (1) by $31$ and the second equation(2) by $28$.
(1)    $31\cdot(28x+17y)\equiv 31\cdot18 \pmod{41}$
(1) $868x+527y\equiv 558\pmod{41}$
(2)   $28\cdot(31x+11y)\equiv 28\cdot35\pmod{41}$
(2) $868x+308y\equiv 980\pmod{41}$
So finally we can subtract equation (1) from (2) we get:
$219y\equiv -422\pmod{41}$
lets check the $\gcd(219,41)$ by Euclidian algorithm :
$219 = 41\cdot 5 + 14$
$41= 14\cdot 2 + 13$
$14= 13\cdot 1 + 1$
$\gcd(219,41)=1$
Hence, because the $\gcd$ is equal to $1$ we can find the Inverse and multiply the equation by the Inverse to find $y$.
$219y\equiv 1\pmod{41}$
$219a = 1+41k$ , the $41k$ must end with the digit of $8$ for $1+$digit $8$ will be $9$ so $k$ must be multiply of number with end digit of $8$.
I don't know how to continue from here .
 [/INST] Alternatively, use linear algebra. The system is
$$
\begin{pmatrix} 28 & 17 \\ 31 & 11 \end{pmatrix} 
\begin{pmatrix} x \\ y\end{pmatrix} 
=
\begin{pmatrix} 18 \\ 35 \end{pmatrix} 
$$
The inverse of the matrix is
$$
-\frac{1}{219}\begin{pmatrix} \hphantom-11 & -17 \\ -31 & \hphantom-28 \end{pmatrix} 
$$
It remains to find the inverse of $219$ mod $41$, which is $3$ by the extended Euclidean algorithm.
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