Shortcut Trick
For infinitely many solutions, the ratios of the coefficients of x, y, and the constant terms must be equal: a1/a2 = b1/b2 = c1/c2.
Substitute values: k/12 = 3/k = −(k − 3)/−k.
Equating the first two ratios: k/12 = 3/k ⇒ k2 = 36 ⇒ k = 6 (since k must be positive).
Verifying with the second and third ratios: 3/6 = (6 − 3)/6 ⇒ 1/2 = 3/6 (True).
∴ The correct answer is 6.
Alternate Method
Given Equations:
Equation 1: kx + 3y − (k − 3) = 0
Equation 2: 12x + ky − k = 0
Comparing with standard form a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0:
a1 = k, b1 = 3, c1 = −(k − 3)
a2 = 12, b2 = k, c2 = −k
Formula Used:
For coincident lines (infinitely many solutions): a1/a2 = b1/b2 = c1/c2
Calculations:
⇒ From the first two ratios: k/12 = 3/k
⇒ k × k = 12 × 3
⇒ k2 = 36
⇒ k = √36 ⇒ k = 6 or k = −6
⇒ Now, check with the second and third ratios: 3/k = (k − 3)/k
⇒ 3k = k(k − 3)
⇒ 3k = k2 − 3k
⇒ k2 − 6k = 0
⇒ k(k − 6) = 0
⇒ k = 0 or k = 6
⇒ The common positive value for both conditions is k = 6.
∴ The correct answer is 6.
Additional Information
Consistent and Inconsistent Systems
A system of equations is consistent if it has at least one solution (unique or infinite) and inconsistent if it has no solution.
Unique Solution (Intersecting Lines)
Condition: a1/a2 ≠ b1/b2. The lines cross at exactly one point (x, y).
No Solution (Parallel Lines)
Condition: a1/a2 = b1/b2 ≠ c1/c2. The lines are parallel and never meet.
Infinite Solutions (Coincident Lines)
Condition: a1/a2 = b1/b2 = c1/c2. Both equations represent the same line.