The question first tells us to build the addition table and multiplication table of congruence class modulo 11.
Now we need to solve X:
$$3x + 5 \equiv 2 \pmod{11}$$
I know how to find $3x \equiv 2 \pmod{11}$ (from the multiplication table) which is $x = 8$ but with $+5$ in the equation I don't know how to do it. Do I just sum it $8 + 5 = 13$?
One could also find some integer K and multiply the whole equation by it. The integer in particular is one such that $3k\equiv 1\bmod 11 $. You can find this by using the Euclidean Algorithm or just by eyeballing it. Let m be the dividend, n be the divisor, and q be the amount of times n goes into m leaving you with a positive remainder r.
$m\;n\;q\;r$ $\quad$This is the equation $m-(n*q)=r$
$11\;3\;3\;2$ $\quad11-(3*3)=2$
$3\;2\;1\;1$ $\quad3-(2*1)=1$
Using the Extended Euclidean Algorithm you can find such a k that works. Let's start with the second line and work our way up. $1=3-(2*1)$ Substituting this into the first line yields $1=3-(1*(11-3*3))=(3*4)-(11*1)$
From here, we have found our k such that $3k\equiv 1 \bmod 11$
If you multiply your original equation by $k=4$, you get $4(3x+5)\equiv (2*4) \bmod 11$
This simplifies down to $12x+20\equiv 8 \bmod 11$. See that $12\equiv 1\bmod 11$
Thus, $x+20\equiv8\bmod 11$.
Adding $-20$ to both sides yields $x\equiv -12\bmod 11$
Now add $22$ to both sides. Note that $22\equiv 0 \bmod 11$
$x+22\equiv 10 \bmod 11$ which simplifies to $x\equiv 10\bmod 11$
Feel free to comment if you are confused about anything. Note: this is one of my first answers so I am sorry about the quality.