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)x – (a + b – 3)y = 4a + b

Answer 1

The given system of equations can be written as

2x – 3y = 7

⇒ 2x – 3y – 7 = 0   ...(i)

And (a + b)x – (a + b – 3)y = 4a + b

⇒ (a + b)x – (a + b – 3)y – 4a + b = 0   ...(ii)

These equations are of the following form:

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

Here, a₁ = 2, b₁ = –3, c₁ = –7 and a₂ = (a + b), b₂ = –(a + b – 3), c₂ = –(4a + b)

For an infinite number of solutions, we must have:

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

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

⇒ `2/(a + b) = 3/((a + b - 3)) = 7/((4a + b))`

⇒ `2/(a + b) = 7/((4a + b))` and `3/((a + b - 3)) = 7/((4a + b))`

⇒ 2(4a + b) = 7(a + b)` and `3(4a + b) = 7(a + b – 3)

⇒ 8a + 2b = 7a + 7b and 12a + 3b = 7a + 7b – 21

⇒ 4a = 17 and 5b = 11

∴ a = 5b   ...(iii)

And 5a = 4b – 21   ...(iv)

On substituting a = 5b in (iv), we get:

25b = 4b – 21

⇒ 21b = –21

⇒ b = –1

On substituting b = –1 in (iii), we get:

a = 5(–1) = –5

∴ a = –5 and b = –1.