Find all solutions $ x $in $Z$ to the simultaneous congruences for $x=1 (mod 2)$ and $x=2 (mod 3)$

Question

In the first part of the question, it asks for 1 solution, so I just made $2k+1=3l+2 $ and found that $k$ is $2$ and $l$ is $1$, so $x=5$. But what does that mean to find "all" solutions for this?

Answer 1

Use Chinese Remainder Theorem to conclude that the answer to the system of equations is $$x = 5 (mod 6)$$

There are more than one values which belong to the above mentioned congruence class. All those are solutions to your system of congruences.

Answer 2

Note that $2|(x+1)$ and $3|(x+1)$.

So $x+1=6k$ ($k\in\Bbb{Z}$) and therefore $$x=6k-1, \;\; k\in\Bbb{Z}.$$