$11x + 13 \equiv 4$ (mod 37) $11x + 13 \equiv 4$ (mod 37)
My "solution" to the problem
$11x + 13 \equiv 4$ (mod 37) $\rightarrow$ 
$11x + 13 = 4 + 37y  $
$11x - 37y = - 9$
Euclid's algorithm.
$37 = 11*3 + 4$
$11 = 4*2 + 3$
$4 = 3*1 + 1 \rightarrow GCD(37,11) = 1$
$3 = 1*3 + 0$
Write as linear equation
$1 = 4 - 1*3$
$3 = 11 - 2*4$
$4 = 37-3*11$
$1 = 4 -1*3 = 4-1(11-2*4) = 3*4-1*11 = 3(37-3*11)-1*11 = 3*37 -10*11$
$1 = 3*37 -10*11$
$11(10) - 37(3) = -1$
$11(90) - 37(27) = -9$
x = 90
That is my answer. But the correct answer should be:
x = 16

We are looking for a solution to $11x\equiv-9\pmod{37}$.
First, solve $11x+37y=1$. As shown in this answer, we use the Extended Euclidean Algorithm:
$$
\begin{array}{r}
&&3&2&1&3\\\hline
1&0&1&-2&3&-11\\
0&1&-3&7&-10&37\\
37&11&4&3&1&0
\end{array}
$$
Thus,
$$
\begin{array}{c}
3\cdot37-10\cdot11=1\\
\Downarrow&\text{multiply by $-9$}\\
90\cdot11\equiv-9\pmod{37}\\
\Downarrow&\text{add $13$ to both sides}\\
16\cdot11+13\equiv4\pmod{37}&90\equiv16\pmod{37}
\end{array}
$$
Therefore, we have
$$
x\equiv16\pmod{37}
$$
