Chinese remainder theorem, can't figure it out! x mod 5 = 3
x mod 7 = 5
x mod 11 = 7
How to determine x? I've been searching on YouTube, but they're giving examples in different ways, for example 
x  ≡  1(mod 3)
I don't understand it, is it the same as x mod 3 = 1?
Thanks, Im still new and quite slow at math, I'm trying to understand it still 

$\,a \equiv b\!\pmod{\! n}\ $ means $\,n\mid a-b,\,$ i.e. $\,a-b = nq\,$ for some integer $q$
$\,a = (b \bmod n)\  $ means the above and $\,0 \le a < n\,$ 
since $ \  b \bmod n\,\ $ denotes the remainder left after dividing $\,b\,$ by $\,n\,$ 

In your system note $\ 3\equiv \color{#c00}{-2}\pmod{\!5},\,\ 5\equiv\color{#c00}{-2}\pmod{\!7},\ $ which yields
$$\begin{align} x\equiv\color{#c00}{-2}\!\!\!\pmod{\!5,7}\iff&\ 5,7\mid x\!+\!2\\[.2em]
 \iff&\ \ \ 35\mid x\!+\!2,\ \ {\rm by}\ \ 35 = {\rm lcm}(5,7)\\[.2em]
 \iff&\ \ \ x\equiv -2 \!\!\pmod{\!35}\end{align}\!$$
Combining the above solution $\bmod 35\,$  with the final congruence   $\ x\equiv\color{#0a0}7\pmod{11}\,$ yields
$35\mid x\!+\!2\,\Rightarrow\, x\!+\!2\bmod 35\cdot 11 = 35\left[\dfrac{x\!+\!2}{35}\bmod{11}\right]\equiv 35[-1]\,$ by $\ \dfrac{\color{#0a0}7\!+\!2}{35}\equiv\dfrac{-2}2\pmod{\!11}$
So we obtain $\ x\!+\!2\equiv -35\,\Rightarrow\, x\equiv -37\equiv 348\pmod{\!385},\ \ 385 = 35\cdot 11$
Remark $ $ Notice that using the relation mod (vs. operation) allowed us to switch from the canonical (least non-negative) remainders $3,5$ to a congruent constant negative residue $\color{#c00}{-2},\,$ so the system simplifies to one where the RHS all have the same value, i.e. $\,x\equiv \color{#c00}{-2}\,$ for both moduli. This constant case system requires only $\rm lcm\,$ (vs. full CRT), i.e. we only need to know that basic fact $\,a,b\mid n\iff {\rm lcm}(a,b)\mid n\,$ in order to solve the equivalent system $\,x\equiv \color{#c00}{-2}\,\pmod{5,7}.\,$ See CCRT = Constant case CRT for more on this (which is equivalent to the uniqueness direction of CRT, i.e. we could have invoked uniqueness vs. using $\rm{lcm}$ above).
