Prove that if a=bmodn and c=dmodn then a+c=b+dmodn and ac=bdmodn

My solution for the first part is:

n|(a-b) so (a-b)=nq and n|(c-d) so (c-d)=np

so (a-b) +(c-d)=nq+np

(a+c)-(b+d)=n(q+p)

So n|[(a+c)-(b+d)]

So

a+c=b+dmodn

I am not sure how to show the second part any help would be great