Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:

(a – 1)x + 3y = 2,


6x + (1 – 2b)y = 6.

Given: (a – 1)x + 3y = 2 – eq 1


6x + (1 – 2b)y = 6 – eq 2


Here,


a1 = (a - 1), b1 = 3, c1 = - 2


a2 = 6 , b2 = (1 - 2b), c2 = - 6


Given that system of equations has infinitely many solution


= =


= =


Here,


=


3×( - 6) = (1 - 2b)×( - 2)


- 18 = - 2 + 4b


4b = - 18 + 2


4b = - 16


b =


b = - 4


Also,


=


- 6(a - 1) = - 2×6


- 6a + 6 = - 12


- 6a = - 12 - 6


- 6a = - 18


a =


a = 3


a = 3


a = 3, b = - 4


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