Solve $x\equiv 1\pmod{5}, x\equiv 2\pmod{6}, x\equiv 3\pmod{7}$ Solve $x\equiv 1\pmod{5}, x\equiv 2\pmod{6}, x\equiv 3\pmod{7}$
First I can see $x=5t+1, t\in Z$. Then they insert this into the second equation, which is $5t+1\equiv 2(mod6)$, which leads to $t\equiv 5mod6$. Then they get $t=6u+5$, I'm confused how they got to $t=6u+5$ from $5t+1$. Also if anyone has any strategies to prove these type of problems where there's an $x$ that has different moduluses.

x % 5 = 1 = -4
x % 6 = 2 = -4
x % 7 = 3 = -4
This means x % LCM(5, 6, 7) = -4
x % 210 = -4
Therefore x = 210t - 4 (-4, 206, ...)
