Show that $R = \{ (x,y) \in \mathbb{Z} \times \mathbb{Z} : 4 \mid(5x+3y)\}$ is an equivalence relation. 
Let $R$ be a relation on $\mathbb{Z}$ defined by $$ R = \{ (x,y) \in \mathbb{Z} \times \mathbb{Z}  :  4 \mid (5x+3y)\}.$$ Show that $R$ is an equivalence relation. 

I'm having a bit of trouble with this exercise in my book and I am trying to study. Can anyone give guidance for this? I know we have to show reflexivity, symmetry, and transitivity, but I don't think what I have on my paper is completely right. I would appreciate other people's opinions on what the solution should be.

Reflexive: $xRx = 5x + 3x = 8x$ which is clearly divisible by $4$ if $x\in\mathbb Z.$
Symmetry: Assume $xRy$, that implies that $5x + 3y$ is divisible by $4$. 
Then $yRx = 5y + 3x = (9y - 4y) + (15x - 12x) = 9y + 15x - 4y - 12x $$= (3y + 5x)- 4(y + 3x).$  Since $3y + 5x$ is divisible by $4$ and $4(y + 3x)$ is divisible by $4$, their difference is divisible by $4$ too, when $x$ and $y$ are integers.
Transitivity: Assume $xRy = 5x + 3y$ and $yRz = 5y + 3z$. Sum them together and you get the $5x + 8y + 3z$, which is divisible by $4$ (since you added two terms divisible by $4$), so $5x + 8y + 3z - 8y$ must also be  divisible by $4$ (because you subtracted a number divisible by $4$ from an expression divisible by $4$) and there you get the $xRz$.
