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

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?

• It should be $3l + 2$. You can yourself verify that your answer 7 does not satisfy the second congruence. Commented Nov 17, 2016 at 1:28
• Yeah you are right, my bad it should have been 3l+2 so x would l=1 and k=2 so x=5. I made a mistake Commented Nov 17, 2016 at 1:33
• $k = 2$ and$l= 1;x = 5$ is one answer. But l =3, k =5; x = 11 is another. and l=5, k = 8; x = 17 is a third. Can you write a formula for all of them? Hint it will be in the form $x = m \mod n$. Commented Nov 17, 2016 at 2:01

Use Chinese Remainder Theorem to conclude that the answer to the system of equations is $$x = 5 (mod 6)$$
Note that $2|(x+1)$ and $3|(x+1)$.
So $x+1=6k$ ($k\in\Bbb{Z}$) and therefore $$x=6k-1, \;\; k\in\Bbb{Z}.$$