$11x + 13 \equiv 4$ (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 have $$11x\equiv -9\equiv 28\mod 37$$ so we get
$$x\equiv \frac{28}{11}\equiv \frac{65}{11}\equiv\frac{102}{11}\equiv \frac{139}{11}\equiv \frac{176}{11}\equiv 16\mod 11$$
