Question

For what value of k does the system of equations

kx + 2y = 5, 3x – 4y = 10

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

Answer 1

The given system of equations:

kx + 2y = 5

⇒ kx + 2y – 5 = 0   ...(i)

3x – 4y = 10

⇒ 3x – 4y – 10 = 0   ...(ii)

These equations are of the forms:

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

where, a₁ = k, b₁ = 2, c₁ = –5 and a₂ = 3, b₂ = –4, c₂ = –10

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

∴ `(a_1)/(a_2) ≠ (b_1)/(b_2)` i.e., `k/3 ≠ 2/(-4) ⇒ k ≠ (-3)/2`

Thus for all real values of k other than `(-3)/2`, 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)`

⇒ `k/3 = 2/(-4) ≠ (-5)/(-10)`

⇒ `k/3 = 2/(-4)` and `k/3 ≠ 1/2`

⇒ `k = (-3)/2, k ≠ 3/2`

Hence, the required value of k is `(-3)/2`.