Question

Find the value of k for which the following system of equations has no solution:

kx + 3y = k – 3

12x + ky = k

Answer 1

Given: kx + 3y = k – 3, 12x + ky = k

Step-wise calculation:

1. Write in the form a₁x + b₁y = c₁ and a₂x + b₂y = c₂: 

a₁ = k, b₁ = 3, c₁ = k – 3

a₂ = 12, b₂ = k, c₂ = k

2. For no solution (parallel, distinct lines) we need

`a_1/a_2 = b_1/b_2 ≠ c_1/c_2`

3. Set `a_1/a_2 = b_1/b_2`: 

`k/12 = 3/k` 

⇒ k² = 36 

⇒ k = ±6

4. Test the two values:

If k = 6: `a_1/a_2 = 6/12 = 1/2` 

`b_1/b_2 = 3/6 = 1/2` 

`c_1/c_2 = (6 - 3)/6 = 3/6 = 1/2` 

All three ratios equal ⇒ infinitely many solutions (coincident lines), not "no solution."

If k = –6: `a_1/a_2 = (-6)/12 = -1/2`

`b_1/b_2 = 3/(-6) = -1/2`

`c_1/c_2 = (-6 - 3)/(-6) = (-9)/(-6) = 3/2`

Here `a_1/a_2 = b_1/b_2` but ≠ `c_1/c_2` ⇒ no solution.

The system has no solution for k = –6.