Question

The number of values of k for which the system of equations (k + 1) x + 8y = 4k, kx + (k + 3) y = 3k – 1 has no solution is-

• A. 0
• B. 1
• C. 2
• D. Infinite

Answer 1

Answer: (B)

1

Answer 2

For the system of equations to have no solution, we must have$\frac{k + 1}{k}$= $\frac{8}{k + 3}$¹ $\frac{4k}{3k - 1}$

Ž (k + 1) (k + 3) = 8k

and 8(3k – 1) ¹ 4k (k + 3) But (k + 1) (k + 3) = 8k

Ž k² + 4k + 3 = 8k

or k² – 4k + 3 = 0 or (k – 1) (k – 3) = 0

Ž k = 1, 3.

For k = 1, 8(3k – 1) = 16 and 4k (k + 3) = 16

\8(3k – 1) = 4k (k + 3) for k = 1 For k = 3, 8(3k – 1) = 64 and 4k (k + 3) = 72

i.e. 8(3k – 1) ¹ 4k (k + 3) for k = 3. Thus, there is just one value for which the system of equations has no solution.