Can someone help me solve this system of congruences? I'm a little new to congruences but I think I have it right.
I started with the following congruences:
$19x \equiv 5 \pmod{2}$ 
$19x \equiv 5 \pmod{3}$ 
$19x \equiv 5 \pmod{5}$ 
$19x \equiv 5 \pmod{7}$
I was able to simplify it to be: 
$x \equiv 1 \pmod{2}$ 
$x \equiv 2 \pmod{3}$ 
$x \equiv 0 \pmod{5}$ 
$x \equiv 1 \pmod{7}$ 
What do I do from here though? I don't know what to do now.
 A: Hint $\,\ 2,3,5,7\mid 19x\!-\!5\iff 210\mid 19x\!-\!5,\,$ since $\,{\rm lcm}(2,3,5,7) = 2\cdot 3\cdot 5\cdot 7 = 210.\,$
${\rm mod}\ 210\!:\ \dfrac{1}{19} \equiv \dfrac{-209}{19}\equiv\dfrac{-190-19}{19}\equiv \color{#c00}{-11}\ $ so $\ 19x\equiv 5\!\!\overset{\ \times\ \color{#c00}{-11}}\iff x\equiv -55\equiv 155$
A: Your reductions are correct, though you shouldn't be including $x \equiv 0 \pmod{5}$. 
We will use the Chinese Remainder Theorem here. The first thing we do is calculate the product of the moduli: $M = 210$. Now we take terms as follows:
Given congruences of the form $x \equiv a_{i} \pmod{m_{i}}$, we generate terms $a_{i} * \frac{M}{m_{i}} * (\frac{M}{m_{i}}^{-1} \pmod{m_{i}})$.
So we have initially $x \equiv 1 \pmod{2}$. So $a_{i} = 1$, $\frac{210}{2} = 105$, and $105^{-1} \equiv 1 \pmod 2$. So our first term is $105$.
For our second term, we look at $2 * 70 * (70^{-1} \pmod{3}) = 140$.
For our third term, we look at $1 * 30 * (30^{-1} \pmod{7}) = 120$.
Now we solve $x \equiv 105 + 140 + 120 \pmod{210}$.
You can find more on the Chinese Remainder Theorem here: http://www.dreamincode.net/forums/topic/238017-chinese-remainder-theorem-tutorial/
