Q:

Proof of: If $a \equiv b \pmod{d}$ and $x \equiv y \pmod{d}$ then $a + x \equiv b + y \pmod{d}$ and $ax \equiv by \pmod{d}$

I am trying to learn mathematics from the beginning, i.e. trying to form a solid foundation and understanding of basic concepts that I should have learned in high school. 
I am working through Basic Mathematics by Serge Lang and I love the book so far.  It is terse and dry in places but I like that, it forces me to really think through things for myself and research other sources when I'm stuck.  
Well, I think I'm somewhat stuck right now.  
I don't know if I understood the exercise well enough and if my proofs are correct. 
If anyone has the book, it is exercise 25, p.26, Algebra, Even and odd integers; divisibility. 
For those who don't own the book:
Let $d$ be a positive integer.  Let $a, b$ be integers.  Define 
$$
a \equiv b \pmod{d}
$$
to mean that $a - b$ is divisible by $d$. Prove that if 
$a \equiv b \pmod{d}$ and $x \equiv y \pmod{d}$,
then
$$a + x \equiv b + y \pmod{d}$$
and
$$ax \equiv by \pmod{d}$$
This was the first time that I met $\equiv$ symbol and $\pmod{d}$. In the book it is only briefly mentioned in the previous exercise (which is the same, just $d = 5$) that we read this "$a$ is congruent to $b$ modulo 5".  I went on to research this a bit since I didn't know how to even start but have found most of the explanations too advanced for my level and I couldn't take much away from it.  After some time I gave up and checked the solution at the back of the book but I didn't understand it quite well.  I took some ideas from it and came up with the following proof, which seems right to me but I would like someone to confirm it since it is somewhat different from the one in the book. 
By definition we have
$$a - b = dk \implies a = b + dk$$
$$x - y = dl \implies x = y + dl$$
for some integers $k, l$.
Proof of the first statement:
$$\begin{align}
a + x & = b + dk + y + dl \\
& = b + y + d(k + l) \\
\end{align}
$$
from which we get
$$(a + x) - (b + y) = d(k + l).$$
Which is the same as saying
$$a + x \equiv b + y \pmod{d}$$
Proof of the second statement:
$$\begin{align}
ax & = (b + dk)(y + dl) \\
& = by + bdl + dky + d^2kl \\
& = by + d(bl + ky + dkl) \\
\end{align}
$$
$$ax - by = d(bl + ky + dkl).$$
Which we can write 
$$ax \equiv by \pmod{d}$$
I think I understood what this means, in a way congruence is similar to equality.  So it is similar to the situation: If $A = B$ and $C = D$ then $A + C = B + D$, and similarly for multiplication. 

A:

Congratulations, your proof is well. 
Regards!