This system of modular equations can be solved by inspection - without rote application of CRT. The first two equations are $\rm\ x \equiv -1\ (mod\ 2,3)\ $ therefore they are equivalent to $\rm\ x\equiv -1\ (mod\ 6)\:.\:$ To solve this and the remaining equation is easy because the moduli $\rm\ m=5\:,\ n=6\ $ are such that one has an obvious inverse modulo the other, viz. $\rm\:6\equiv 1\ (mod\ 5)\:$ hence $\rm\:6^{-1}\equiv 1^{-1}\ \equiv 1\ (mod\ 5)\:.\:$ Therefore, applying EasyCRT below with $\rm\:b = -1\:$ we conclude $\rm\ x\ \equiv\ b + n\ \bigg[\frac{a-b}{n}\ mod\ m\bigg]\ \equiv\ -1 + 6\ (a+1)\ \equiv\ 5+6a\ \ (mod\ 30)\:.$
These special cases of CRT, where the RHS is constant, and where one modulus has obvious inverse mod the other, are well worth knowing, since they arise frequently in practice, so they often go a long way towards shortening manual calculations. Below is the EasyCRT that I employ to easily compute the solution when one modulus $\rm\:n\:$ has obvious inverse mod the other $\rm\:m\:.$
THEOREM (EasyCRT) $\rm\ \ $ If $\rm\ m,\:n\:$ are coprime integers then $\rm\ n^{-1}\ $ exists $\rm\ (mod\ m)\ \ $ and
$\rm\displaystyle\quad\quad\quad\quad\quad x\equiv a\ (mod\ m),\ \ x\equiv b\ (mod\ n)\ \ \iff\ \ x\ \equiv\ b + n\ \bigg[\frac{a-b}{n}\ mod\ m\:\bigg]\ \ (mod\ m\:n)$
Proof $\rm\ (\Leftarrow)\ \ \ mod\ n\::\ x\equiv b + n\ (\cdots)\equiv b\:,\ $ and $\rm\ mod\ m\::\ x\equiv b + (a-b)\ n/n \equiv a\:.$
$\rm\ (\Rightarrow)\ \ $ The solution is unique $\rm\ (mod\ m\:n)\ $ since if \rm\ x',\:x\ are solutions then \rm\ x'\equiv x\ mod $\rm\:m,n\:$ therefore \rm\ m,\:n\ |\ x'-x\ \Rightarrow\ m\:n\ |\ x'-x\ \ since $\rm\ \:m,\:n\:$ coprime $\rm\:\Rightarrow\ lcm(m,n) = m\:n\:.\quad\quad$ QED