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.

 A: From $1=(4)25 - (11)9$ you get $5 = (20)25-(55)9$, not $5=55(9) - (20)25$, which has the wrong sign. So $x \equiv -55 \equiv -5 \equiv 20 \bmod 25$.
A: $$\text{Hint: Use euclidean algorithm}$$
$$\text{Find $a,b$ such that $9a+25b=1$, which can be done since gcd(9,25)=1}$$
$$\text{You'll finish with: $1 \equiv 9 \cdot a \mod (25) \Rightarrow x \equiv 9x \cdot a (\mod 25)$. Remember $9x \equiv 5 (\mod 25)$}$$
A: There is a much better way to look at this.
The whole idea of modulos is that $9x=5+25 \cdot n$ for any integer $n$.
What you can do then is keep adding 25 to the right hand side until it is divisible by 9. As it turns out 180 is the first such number so we have 
$$9x \equiv 180 \pmod{25}$$ as gcd(9,25)=1 we can divide by 9 to get
$$x \equiv 20 \pmod{25}$$
You only have to sub your answer back in to see that it's wrong.
A: 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.
