Check:

(a,b)R (a,b) ∈ R

a + b = b + a

hence R is reflexive.

Now,

(a, b) R (c, d) ∈ R

a + d = b + c

⇒ c + b = d + a

⇒ (c,d) R (a,b) ∈ R

R is symmetric

Now,

(a, b) R (c, d) ∈ R

a + d = b + c

(c,d) R (e,f) ∈ R

⇒ c + f = d + e

Now,

a + d + c + f = b + c + d + e

⇒ a + f = b + e

So (a,b) R (e,f)

R is transitive.

Hence R is an equivalence relation.