system of linear congruences How can I solve this system of linear congruences? $$2x\equiv 0\, \text{mod 3}\\ 3x\equiv 2\, \text{mod 5}\\ 5x\equiv 4\, \text{mod 7}$$
I don´t know where to start, I am having a lot of troubles, can you lend me a hand please?

Let's solve this equation without using the CRT. so $2x = 3k$, and this requires that $ 2 | k$. So $k = 2t$. So $x = 3t$, and $3(3t) = 5m + 2$. So $9t = 5m + 2 = 9m + 2 - 4m$, and $t = m + \dfrac{2 - 4m}{9}$. So $2 - 4m = 9p$, and $m = \dfrac{2 -9p}{4} = -2p + \dfrac{2 - p}{4}$. So $p = 2 - 4q$. So $t = -2(2 - 4q) + q + 2 - 4q = -2 + 5q$. So $x = 3(-2 + 5q) = -6 + 15q$. So $5(-6 + 15q) = 7w + 4$ gives: $-30 + 75q = 7w + 4$ or $75q = 7w + 34 = 75w + 34 - 68w = 75w + 34(1 - 2w)$. So $q = w + \dfrac{34(1 - 2w)}{75}$. So $1 - 2w = 75u$. So $w = \dfrac{1 - 75u}{2} = -38u + \dfrac{u+1}{2}$. So $u = 2z - 1$. So $w = -38(2z - 1) + z = -75z + 38$. So $q = -75z + 38 + 34(2z - 1) = -7z + 4$. So $x = -6 + 15q = -6 + 15(-7z + 4) = -105z + 54$, with $z \in \mathbb{Z}$. So the answer is: $x = -105z + 54$, with $z \in \mathbb{Z}$. Check each equation holds.
