a Ra⇒5a is multiple it 5 So reflexive aRb⇒2a+3b=5α, Now b R a 2b+3a‌=2b+(‌
5α−3b
2
)⋅3 ‌=‌
15
2
α−‌
5
2
b=‌
5
2
(3α−b) ‌=‌
5
2
(2a+2b−2α) ‌=5(a+b−α) Hence symmetric ‌‌ a R b ‌‌‌⇒2a+3b=5α‌. ‌ ‌‌ b R c ‌‌‌⇒2b+3c=5β ‌‌ Now ‌‌‌2a+5b+3c=5(α+β) ‌⇒2a+5b+3c=5(α+β) ‌⇒2a+3c=5(α+β−b) ‌⇒aRc ‌ Hence relation is equivalence relation.