Question

For what value of k does the system of equations

x + 2y = 5, 3x + ky + 15 = 0

have (i) a unique solution, (ii) no solution?

Answer 1

The given system of equations:

x + 2y = 5

⇒ x + 2y – 5 = 0   ...(i)

3x + ky + 15 = 0   ...(ii) 

These equations are of the forms:

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

where, a₁ = 1, b₁ = 2, c₁ = –5 and a₂ = 3, b₂ = k, c₂ = 15

(i) For a unique solution, we must have:

∴ `(a_1)/(a_2) ≠ (b_1)/(b_2)` i.e., `1/3 ≠ 2/k ⇒ k ≠ 6`

Thus, for all real values of k other than 6, the given system of equations will have a unique solution.

(ii) For the given system of equations to have no solutions, we must have:

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

⇒ `1/3 = 2/k ≠ (-5)/15`

⇒ `1/3 = 2/k` and `2/k ≠ (-5)/15`

⇒ k = 6, k ≠ –6

Hence, the required value of k is 6.