# Solve $3y + 2 \equiv 3\ (5)$ in math exercise

I'm stuck at the ending part of a math exercise on congruences.

I must solve the following system of congruences $S$:

$x \equiv 2\ (3)$

$x \equiv 3\ (5)$

I was first asked to give the remainders of the division of $3y +2$ by 5, with knowing the remainders of the division of $y$ by 5.

Here's what I did:

-If $y \equiv 0\ (5)$, then $3y +2 \equiv 2\ (5)$

-If $y \equiv 1\ (5)$, then $3y +2 \equiv 0\ (5)$

-If $y \equiv 2\ (5)$, then $3y +2 \equiv 3\ (5)$

-If $y \equiv 3\ (5)$, then $3y +2 \equiv 1\ (5)$

-If $y \equiv 4\ (5)$, then $3y +2 \equiv 4\ (5)$

Here's the part of my exercise I'm stuck with:

I must find the solutions of $3y +2 \equiv 3\ (5)$, knowing that each solution $x$ is like $x$ = $15z + 8$ with $z$ which is an integer. Then, I will need to prove that each integer $x$ like $x = 15z + 8$ is a solution of the system $S$.

I really don't know what to do with this part of my exercise, what shall I do?

Hint : $15z+8\equiv 8\ (\ mod\ 3\ )$ and $15z+8\equiv 8\ (\ mod\ 5\ )$ because of $3|15$ and $5|15$.
• Note that $8$ is a solution, hence $15z+8$ is a solution for every $z\in\mathbb Z$ – Peter Dec 2 '15 at 14:18