Find all solutions of this system: x is congruent to 2 (mod 3), 2x is congruent to 3 (mod 5) and 3x is congruent to 4 (mod 7).

It is possible to solve this system using the
Linear Congruence Theorem and the Extended Euclidean
Algorithm.

We must solve the following system of
linear congruences:

`x -= 2 (mod
3)`

`2x -= 3 (mod 5)`

`3x -= 4
(mod 7)`

Conveniently, the first congruence is already
written in terms of x. We can alternatively write this as x = 3k + 2.
We will substitute of x into our next
equation:

`2x -= 3 (mod 5) implies 2(3k +2) -= 3(mod 5)
implies 6k -= -1 (mod 5)`

Then solving for k using the
technique described on the Linear congruence theorem reference page, we find ` k -= 4
(mod 5)` , or k = 5l + 4

Plugging this
back into our equation for x, we find

`x = 3k + 2 = 3(5l +
4) + 2 = 15l + 12 + 2= 15l + 14.`

We can then plug this
into our last equation.

`3x -= 4 (mod 7) implies 3(15l +
14) -= 4(mod 7) implies 45l -= -38 (mod 7)`

Then solving
for l, we find ` l -= 1 (mod 7)` , or ` l = 7m + 1`
.

But we know x = 5l + 5, so plugging this back in we find
that

`x = 5(7m + 1) + 5 = 35m + 5 + 5 = 35m + 10`
.

Therefore, the solution to the system is `x
-= 10 (mod 35)` .