Find the least positive integer $x$ which satisfies the following simultaneous equations:
$ 2x ≡ 1 \pmod 5 $
$ 3x ≡ 9 \pmod 6 $
$ 4x ≡ 1 \pmod 7 $
$ 5x ≡ 9 \pmod {11} $
We can rewrite the equations to isolate the variable $x$
$ 2x ≡ 1 \pmod 5 \iff 6x \equiv 3 \pmod 5 \iff x \equiv 3 \pmod 5$
$ 3x ≡ 9 \pmod 6 \iff 3x \equiv 3 \pmod 6 \iff x \equiv 1 \pmod {\frac{6}{\gcd(6,3)}} \iff x \equiv 1 \pmod 2 $
$ 4x ≡ 1 \pmod 7 \iff 8x \equiv 2 \pmod 7 \iff x \equiv 2 \pmod 7 $
$ 5x ≡ 9 \pmod {11} \iff 45x \equiv 81 \pmod {11} \iff x \equiv 4 \pmod {11} $
The modulii 5, 2, 7 and 11 are pair-wise coprime, and so we can use the Chinese Remainer Theorem to solve the following simultaneous linear congruences.
$ x \equiv 3 \pmod 5$
$ x \equiv 1 \pmod 2 $
$ x \equiv 2 \pmod 7 $
$ x \equiv 4 \pmod {11} $
Formula (3.23) here is
$$ x = a_1 N_1 x_1 + a_2 N_2 x_2 + a_3 N_3 x_3 + a_4 N_4 x_4 = 3(2 \times 7 \times 11)x_1 + 1(5 \times 7 \times 11)x_2 + 2(5 \times 2 \times 11)x_3 + 4(5 \times 2 \times 7)x_4 $$
where $N_k x_k \equiv 1 \pmod {n_k}$, and so
$$ \begin{align} 154 x_1 & \equiv 1 \pmod 5 \\ \\ 4 x_1 & \equiv 1 \pmod 5 \\ \\ x_1 & \equiv 4 \pmod 5 \end{align}$$
and
$$ \begin{align} 385 x_2 & \equiv 1 \pmod 2 \\ \\ x_2 & \equiv 1 \pmod 2 \end{align}$$
also
$$ \begin{align} 110 x_3 & \equiv 1 \pmod 7 \\ \\ 5 x_3 & \equiv 1 \pmod 7 \\ \\ x_3 & \equiv 3 \pmod 7 \end{align}$$
and also
$$ \begin{align} 70 x_4 & \equiv 1 \pmod {11} \\ \\ 4 x_4 & \equiv 1 \pmod {11} \\ \\ x_4 & \equiv 3 \pmod {11} \end{align}$$
This gives
$$ x = a_1 N_1 x_1 + a_2 N_2 x_2 + a_3 N_3 x_3 + a_4 N_4 x_4 = 3(2 \times 7 \times 11)4 + 1(5 \times 7 \times 11)1 + 2(5 \times 2 \times 11)3 + 4(5 \times 2 \times 7)3 = 3733 $$
The general unique solution is
$$ \begin{align} x & \equiv 3733 \pmod {5 \times 2 \times 7 \times 11} \\ \\ x & \equiv 653 \pmod {770} \end{align}$$
That is, for some integer $t$
$$ x = 653 + 770t$$
The least positive solution is $653$.