Prove that $\forall x,y\in\mathbb{Z},x + 3y$ is a multiple of $7$ iff $3x + 2y$ is a multiple of $7$

Question

I'm struggling with this problem, could someone point me in the right direction?

Prove that for all $x$ and $y$ in set $\mathbb{Z}$, $x+3y$ is a multiple of $7$ iff $(3x + 2y)$ is a multiple of $7$

I don't really want the answer just some direction on how to work through this. Thanks.

Answer 1

$$ ( x+3y) + 2(3x+2y) = 7(x+y)$$

Thus $(x+3y)$ is a multiple of $7$ iff $2(3x+2y)$ is a multiple of $7$

Since $2$ and $7$ are relatively prime, the above statement is equivalent to $(x+3y)$ is a multiple of $7$ iff $(3x+2y)$ is a multiple of $7$

Answer 2

The idea is to eliminate one of the variables.

$$3x+2y-3(x+3y)=-7y$$

So, $7|(3x+2y)\iff7|3(x+3y)$

But $7\nmid 3,7|3(x+3y)\implies7|(x+3y)$

Alternatively, $$3(3x+2y)-2(x+3y)=7x$$

So, $7|3(3x+2y)\iff7|2(x+3y)$

But $7\nmid 3,7\nmid2$

Answer 3

$$\begin{array} {rrll} & 0 \equiv & x + 3y \\ \iff & 0 \equiv & 3(x + 3y) \\ \iff & 0 \equiv & 3x + 2y & \pmod 7 \end{array}$$

Answer 4

Prove: $$7\mid x + 3y \iff 7\mid 3x + 2y$$

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Say $7\mid x+3y $ then $x+3y = 7k$ so $x= 7k-3y$, so we have $$3x+2y = 3(7k-3y)+2y =21k-7y = 7(3k-y) \implies 7\mid 3x+2y$$

Say $7\mid 3x+2y$ then $3x+2y = 7l$ so $y = {7l-3x\over 2}$, so we have $$ x+3y = {2x+21l-9x\over 2} = 7{3l-x\over 2} \implies 7\mid 2(x+3y) \implies 7\mid x+3y$$