Given: 2x - 3y = 7 – eq 1

(a + b)x - (a + b - 3)y = 4a + b – eq 2

Here,

a₁ = 2, b₁ = - 3, c₁ = - 7

a₂ = (a + b), b₂ = - (a + b - 3), c₂ = - (4a + b)

Given that system of equations has infinitely many solution

∴ = =

= =

Here,

=

- 3×( - 4a + b) = - 7× - (a + b - 3)

12a + 3b = 7a + 7b - 21

12a - 7a = - 3b + 7b - 21

5a = 4b - 21

5a – 4b + 21 = 0 eq 3

Also,

=

2× - (4a + b) = - 7×(a + b)

- 8a – 2b = - 7a – 7b

- 8a + 7a = 2b – 7b

- a = - 5b

a = 5b eq 4

substitute – eq 4 in – eq 3

5(5b) – 4b + 21 = 0

25b – 4b + 21 = 0

21b + 21 = 0

b =

b = - 1

substitute ‘b’ in – eq 4

a = 5( - 1)

a = - 5

∴ a = - 5, b = - 1