Chinese remainder theorem issue Let's say I have the following equations:
$$x \equiv 2 \mod 3$$
$$x \equiv 7 \mod 10$$
$$x \equiv 10 \mod 11$$
$$x \equiv 1 \mod 7$$
And I need to find the smallest x for which all these equations are correct. So:
$N = 3\times10\times11\times7 = 2310$
$N_1 = \frac{2310}{3} = 770$
$N_2 = \frac{2310}{10} = 231$
$N_3 = \frac{2310}{11} = 210$
$N_4 = \frac{2310}{7} = 330$
Then:
$\gcd (3, 770) = 1 = 257\times3 - 1\times770; \qquad x_{1} = -1 \equiv 2 \mod 3$
$\gcd(10, 231) = 1 = -23\times10 + 1\times231; \quad x_{1} = 1$
Etc., and finally:
$x = 2\times2\times N_1 + 7\times1\times N_2 + ...$

I understand the whole method except for one thing - when I solved $\gcd(3, 770)$ I got $x_{1} = -1$...why is there $-1 \equiv 2 \mod 3$, from which the final result is 2?

More simply: $ $ mod $3,11\!:\ x\equiv -1,\ $ so $ $ mod $33\!:\ x\equiv -1\ $ so $\ x = -1 + 33j.$
mod $7\!:\ 1 \equiv x \equiv -1 + 33j\ \Rightarrow\ j\equiv \frac{2}{33} \equiv \frac{2}{{-}2} \equiv -1\ $ so $\ j = -1 + 7k.$  
Substituting:  $\ x = -1 + 33j = -1 + 33(-1 + 7k) = -34 + 231k.$
mod $10\!:\ 7 \equiv x \equiv -34 + 231k \equiv -4 + k\ $ so $\ k\equiv 7+4\equiv 1,\ $ i.e. $\ k = 1+10n.$
Substituting: $\ x = -34 + 231k = -34 + 231(1+10n) = 197 + 2310n.$
