How can i show that $b-a = d-c$ is transitive?

Question

Let R be a binary relation on the set of ordered pairs of integers such that $R={(a,b),(c,d))| b-a=d-c} $. Show that R is an equivalent relation.

What I got so far is this, i'm not sure if its right, but I'm struggeling with showing that R is transitive.

PROOF: Reflexive: Let

$(a,b)∈A$

Since $b-a$ can be written as $b+(-a)$. We get

$b-a=(-a)+b$

$((a,b),(a,b))∈R$

Thus R is reflexive.

Symmetric: Let

$ ((a,b),(c,d))∈R$

$b-a=d-c$

Use the commutative property for adding, since both sides can be written as shown over. We get

$(-a)+b=(-c)+d$

Which is equivalent with : $(-c)+d= (-a)+b$

Which implies

$((c,d),(a,b))∈R$

Thus R is symmetric.

Transitive Let $((a,b),(c,d))∈R$ and $((c,d),(e,f))∈R$

Answer 1

You know that $b-a=d-c$. And you know that $d-c=f-e$. Therefore, $b-a=f-e$, which means that $\bigl((a,b),(e,f)\bigr)\in R$.