Given: 2x + 3y = 7 – eq 1

(a + b + 1)x + (a + 2b + 2)y = 4(a + b) + 1 – eq 2

Here,

a₁ = 2, b₁ = 3, c₁ = - 7

a₂ = (a + b + 1), b₂ = (a + 2b + 2), c₂ = - (4(a + b) + 1)

Given that system of equations has infinitely many solution

∴ = =

= =

Here,

=

3× - (4(a + b) + 1) = - 7×(a + 2b + 2)

- 12a - 12b - 3 = - 7a - 14b - 14

- 12a + 7a - 3 = 12b - 14b - 14

- 5a - 3 = - 2b - 14

5a - 2b - 11 = 0 eq 3

Also,

=

2× - (4(a + b) + 1) = - 7×(a + b + 1)

- 8a – 8b – 2 = - 7a – 7b – 7

- 8a + 7a = 8b – 7b – 7 + 2

- a = b – 5

a + b = 5

a = 5 – b eq 4

substitute – eq 4 in – eq 3

5(5 – b) - 2b - 11 = 0

25 – 5b - 2b - 11 = 0

- 7b + 14 = 0

b =

b = 2

substitute ‘b’ in – eq 4

a = 5 - 2

a = 3

∴a = 3, b = 2