Calculus showing with mods

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I got this problem from my Calculus II teacher, I have no idea how to approach it...

Show that if $a,b,c$, and $d$ are integers such that $a\equiv b\pmod m$ and $c\equiv d\pmod m$, then $a+c\equiv b+d\pmod m$ and $ac\equiv bd\pmod m$.

If anyone can help, thanks!

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If $a \equiv b\ \ \text{mod}\ m$ and $c \equiv d\ \ \text{mod}\ m$ it follows* that $m |(a-b)$ and $m|(c-d)$ then there exist $r_1,r_2 \in \mathbb{Z}$ such that $$a-b = m\ r_1\ \ \text{and}\ \ c-d = m\ r_2$$ What can you say about $(a-b) + (c-d)$?

You take from here.

Same idea goes for $ac \equiv bd\ \ \text{mod}\ m$.

(*) Show that $a \equiv b\ \ \text{mod}\ \ m \Leftrightarrow m|(a-b)$ (It folllows directly from two facts : (i) Euclidean Division (ii) if $x|(y+z)$ and $x|y$ then $x|z$).