Answer :
Proof :
First, it is important to have in mind that a number [tex] m \in \mathbb{Z} [/tex] is a multiple of [tex]n\in\mathbb{Z} [/tex] iff there exists [tex]k\in\mathbb{Z}[/tex] such that [tex] m = n \cdot k[/tex].
Also, you have to prove a logical equivalence. To this end, it is possible to prove two logical implications.
Step-by-step explanation:
1.) Let x, y be integers such that x + 3y is a multiple of 7. You have to prove that 3x +2y is a multiple of 7.
In effect, by hypothesis there exists k [tex]\in\mathbb{Z}[/tex] such that x + 3y = 7 k . So, you get
[tex]\begin{equation*} 4(x+3y) + (3x + 2y) = 7x + 14y = 7 (x + 2y) \ \mbox{(direct computations and factoring)}\end{equation*} [/tex].
Therefore, 4(x +3y) + (3x +2y) is a multiple of 7. Then,
[tex](3x + 2y) = 7 (x + 2y) - 4(x + 3y) = 7 (x+2y) - 4 \cdot 7 k = 7 (x + 2y -4k) \ \mbox{(factoring)}[/tex].
Given that x,y,k are integers, then x + 2y - 4k is an integer and hence, 3x + 2y is a multiple of 7.
To finish, it remains to prove its reciprocal statement.
2.) Let x, y be integers such that 3x + 2y is a multiple of 7. You have to prove that x +3y is a multiple of 7. Reasoining as before , there exists q [tex]\in\mathbb{Z}[/tex] such that 3x + 2y = 7 \cdot q. Thus,
[tex]$ \begin{equation*} 2(3x+2y) + (x + 3y) = 7x + 7y = 7 (x + y) \ \mbox{direct computations and factoring} \\\end{equation*} $[/tex] Thus, [tex] 2(3x +2y) + (x +3y)[/tex] is a multiple of 7.
On the other hand, using the hypothesis [tex] $ \begin{equation*} (x + 3y) = 7 (x + y) - 2(3x + 2y) = 7 (x+y) - 2 \cdot 7 q = 7 (x + y -2q) \ \mbox{(factoring)} \end{equation*} $ [/tex] .
Finally, thanks that [tex]x,y,q [/tex] are integer numbers, then [tex] x + y - 2q[/tex] is a integer number and therefore, [tex] 3x + 2y [/tex] is a multiple of 7.