Question

For what value of k does the system of equations

x + 2y = 3, 5x + ky + 7 = 0

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

Also, show that there is no value of k for which the given system of equations has infinitely many solutions.

Answer 1

The given system of equations:

x + 2y = 3

⇒ x + 2y – 3 = 0   ...(i)

And 5x + ky + 7 = 0   ...(ii)

These equations are of the following form:

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

where, a₁ = 1, b₁ = 2, c₁ = –3 and a₂ = 5, b₂ = k, c₂ = 7

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

∴ `(a_1)/(a_2) ≠ (b_1)/(b_2)` i.e., `1/5 ≠ 2/k ⇒ k ≠ 10`

Thus, for all real values of k other than 10, the given system of equations will have a unique solution.

(ii) In order that the given system of equations has no solution, we must have:

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

⇒ `1/5 ≠ 2/k ≠ (-3)/7`

⇒ `1/5 ≠ 2/k` and `2/k ≠ (-3)/7`

⇒ `k = 10, k ≠ 14/(-3)`

Hence, the required value of k is 10.

There is no value of k for which the given system of equations has an infinite number of solutions.