$x$ is odd if and only if $3x+6$ is odd 
Prove the following proposition. Let $x\in\Bbb Z$. Then $x$ is odd if and only if $3x+6$ is odd. 

I'm currently not seeing a way to transform $3x+6$ into the format of $2k+1$ in order to prove odd. This is my first time dealing with discrete mathematics and proofs and I'm trying to get a feel for it.

Hint: Write $3x+6=x+2(x+3)$  .
