Where's my error on finding all the solutions of a linear congruence? I'm supposed to find all solutions of each of the linear congruence.
9x ≡ 5 (mod 25)

I know there are other posts on the site about this, but I don't really follow.

Here's what I did:
I used the Euclidean Algorithm to find the gcd, which was 1 and then to find the equation, I ended up with
1=(4)25 - (11)9
Then I multiplied by 5 on both sides to get it in the form of the original and got:
5=(20)25 - (55)9
Then 55(9) - 25(20) = 5
So I had x ≡ 55 (mod 25) or x ≡ 5 (mod 25).
But the book had x ≡ 20 (mod 25)
What did I do wrong?

Here's my exact work using Euclidean:  My sign seems correct though.


You incorrectly flipped a sign between two lines. It should be $\ 5 = \color{#c00}{9(-55)} + 20(25)\ $ therefore $\ {\rm mod}\ 25\!:\ 9x\equiv 5 \equiv \color{#c00}{9(-55)},\,$ so $\,x\equiv -55\equiv -5\equiv 20,\,$ by canceling $\,9,\,$ valid by $\,9\,$ is coprime to $25.$ 
Alternatively $\ {\rm mod}\ 25\!:\ x\equiv \dfrac{5}{9}\equiv \dfrac{30}9\equiv \dfrac{10}3\equiv \dfrac{-15}3\equiv -5,\ $ valid since all denominators are coprime to $25$ so invertible.
