Given: 2x + 3y = 7 – eq 1

(a + b)x + (2a - b)y = 21 – eq 2

Here,

a₁ = 2, b₁ = 3, c₁ = - 7

a₂ = (a + b), b₂ = (2a – b), c₂ = - 21

Given that system of equations has infinitely many solution

∴ = =

= =

Here,

=

3× - 21 = - 7×(2a - b)

- 63 = - 14a + 7b

14a - 7b - 63 = 0

2a – b – 9 = 0 eq 3

Also,

=

2× - 21 = - 7×(a + b)

- 42 = - 7a – 7b

7a + 7b + 42 = 0

a + b + 6 = 0

a + b = 6

a = 6 – b eq 4

substitute – eq 4 in – eq 3

2(6 – b) – b – 9 = 0

12 – 2b – b – 9 = 0

- 3b + 3 = 0

b =

b = 1

substitute ‘b’ in – eq 4

a = 6 - 1

a = 5

∴ a = 5, b = 1