For infinitely many solution

⇒ 6a – 3 = 15 and -9 = 5b – 10

⇒ a = 3 and b = 1/5

(ii) For infinitely many solution

⇒ 6 = 2b + 1 and 6a + 15 = 9

⇒ b = 5/2 and a = -1

(iii) For infinitely many solution

⇒ 3a – 3 = 6 and 9 = 1 – 2b

⇒ a = 3 and b = -4

(iv) For infinitely many solution

⇒ 15a – 3 = 12a + 12b and 20a – 4 = 24a – 24b

⇒ 3a – 12b = 3 ------ (1) and 6b – a = 1 ------ (2)

Multiplying eq2 by 3 and adding to eq1

⇒ 6b = 6

⇒ b = 1

Thus, 3a – 12 = 3

⇒ a = 5

(v) For infinitely many solution

⇒ 2a + 2b = 3a – 3b and 6a + 2b – 4 = 7a – 7b

⇒ a = 5b and 9b – a = 4

Thus, 9b – 5b = 4

⇒ b = 1

⇒ a = 5

(vi) For infinitely many solution

⇒ 2a + 2 = 3a – 3

⇒ a = 5

(vii) For infinitely many solution

⇒ 2a + 4 = 3a – 3

⇒ a = 7