Question

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

2x + 3y = 7, (a + b + 1)x – (a + 2b + 2)y = 4(a + b) + 1

Answer 1

The given system of equations can be written as

2x + 3y = 7

⇒ 2x + 3y – 7 = 0   ...(i)

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

(a + b + 1)x – (a + 2b + 2)y – [4(a + b) + 1] = 0   ...(ii)

These equations are of the following form:

a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0

where, a₁ = 2, b₁ = 3, c₁ = –7 and a₂ = (a + b + 1), b₂ = (a + 2b + 2), c₂ = –[4(a + b) + 1]

For an infinite number of solutions, we must have:

`(a_1)/(a_2) = (b_1)/(b_2) = (c_1)/(c_2)`

`2/((a + b + 1)) = 3/((a + 2b + 2)) = (-7)/(-[4(a + b) + 1])`

⇒ `2/((a + b + 1)) = 3/((a + 2b + 2)) = 7/([4(a + b) + 1])`

⇒ `2/((a + b + 1)) = 3/((a + 2b + 2))` and `3/((a + 2b + 2)) = 7/([4(a + b) + 1])`

⇒ 2(a + 2b + 2) = 3(a + b + 1) and 3[4(a + b) + 1] = 7(a + 2b + 2)

⇒ 2a + 4b + 4 = 3a + 3b + 3 and 3(4a + 4b + 1) = 7a + 14b + 14

⇒ a – b – 1 = 0 and 12a + 12b + 3 = 7a + 14b + 14

⇒ a – b = 1 and 5a – 2b = 11

a = (b + 1)   ...(iii)

5a – 2b = 11   ...(iv)

On substituting a = (b + 1) in (iv), we get:

5(b + 1) – 2b = 11

⇒ 5b + 5 – 2b = 11

⇒ 3b = 6

⇒ b = 2

On substituting b = 2 in (iii), we get:

a = 3

∴ a = 3 and b = 2.