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