Question

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

kx + 3y = k – 3 

12x + ky = 6

Answer 1

Given: The system kx + 3y = k – 3, 12x + ky = 6.

Step-wise calculation:

1. Rewrite in standard form:

kx + 3y – (k – 3) = 0 and 12x + ky – 6 = 0

So a₁ = k, b₁ = 3, c₁ = –k + 3

a₂ = 12, b₂ = k, c₂ = –6

2. For no solution (parallel distinct lines) require `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. Check c-ratio:

If k = 6: `a_1/a_2 = b_1/b_2 = 1/2` 

And `c_1/c_2 = (-6 + 3)/(−6) = (-3)/(-6) = 1/2` 

⇒ ratios equal

⇒ lines coincide (infinitely many solutions)

If k = –6: `a_1/a_2 = b_1/b_2 = -1/2` 

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

⇒ c-ratio ≠ a-ratio

⇒ parallel distinct

⇒ no solution

The system has no solution exactly when k = –6.