Answer :

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


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