I'm trying to solve the following set of congruence equations:
$$\left\{ \begin{gathered} x \equiv 1(\bmod 7) \hfill \\ x \equiv 6(\bmod 11) \hfill \\ x \equiv 5(\bmod 13) \hfill \\ \end{gathered} \right.$$
First, we can see that the 1st equation implies $x = 1 + 7z,z \in \mathbb{Z}$, so I substitute it instead of $x$ in the 2nd equation, i.e.
$$1 + 7z = 6(\bmod 11)\quad \Rightarrow \quad 7z = 5 + 11t,t \in \mathbb{Z}\quad \Rightarrow \quad 6 + 11t$$
But, already at this stage, it is evident that if we pick a number $t\in\mathbb{Z}$, the equation $6+11t$ will not equal to $1$ modulo $7$, eg. $6+11=17\quad\implies\quad 17 \mod 7 =3$.
Why is this case? Is it important to try to reduce the equation to a single term, i.e. $z=\dots$?, vs. the way I've done?
You left off the "$z=$" in "$\implies 6+11t$". If you let $z=6+11t$ then $x=1+7z = 1+7(6+11t)$ and that will work just fine.