<s>[INST] Solving simultaneous congruences Trying to figure out how to solve linear congruence by following through the sample solution to the following problem:

$x \equiv 3$ (mod $7$) $x \equiv 2$ (mod $5$) $x \equiv 1$
  (mod $3$)

Let:
$n_1$ = 7 
$n_2 = 5$
$n_3 = 3$
$N = n_1 \cdot n_2 \cdot n_3 = 105$
$m_1 = \frac{N}{n_1} = 15$
$m_2 = \frac{N}{n_2} = 21$
$m_3 = \frac{N}{n_3} = 35$
$gcd(m_1,n_1)$ = $gcd(15,7) = 1 = 15 \times 1 - 7 \times 2$ so $y_1 = 1$ and $x_1 = 15$
$gcd(m_2,n_2)$ = $gcd(21,5) = 1 = 21 \times 1 - 5 \times 4$ so $y_2 = 1$ and $x_2 = 21$
$gcd(m_3,n_3)$ = $gcd(35,3) = 1 = -35 \times 1 + 3 \times 12$ so $y_3 = -1$ and $x_3 = -35$
I understand up to this point, but the next line I don't get:
So $x = 15 \times 3 + 21 \times 2 - 35 \times 1 \equiv 52$ (mod $105$)
Where is the $\times 3$, $\times 2$, $\times 1$ from? Is it just because there are 3 terms, so it starts from 3 then 2 then 1? And where is the 52 coming from?
 [/INST] $ 2x \equiv  -1\,$ mod $\,3,5,7\ \begin{array}{r}\iff 3,5,7\mid 2x\!+\!1\!\\ \iff\ \ \ 105\mid 2x\!+\!1\end{array}\!\iff\!$ mod $\,105\!:\  2x \equiv -1\equiv 104\!\iff\! x\equiv 52\ 
 $
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