For which value of k do the two linear equations kx + 3y = k − 3 and 12x + ky = k have infinitely many solutions?

Difficulty: Medium

0
-6
6
1
3

Correct Answer: 6

Explanation:

Introduction:
Two linear equations in two variables have infinitely many solutions when they are proportional (i.e., represent the same line). Coefficients of x, y, and constants must be in the same ratio.

Given Data / Assumptions:

Equation 1: kx + 3y = k − 3.
Equation 2: 12x + ky = k.
We seek k such that both equations are scalar multiples.

Concept / Approach:
For ax + by = c and a′x + b′y = c′ to be multiples: a/a′ = b/b′ = c/c′. Apply to the given coefficients and solve for k, taking care that denominators are nonzero.

Step-by-Step Solution:

Compare ratios: k/12 = 3/k = (k − 3)/k.From k/12 = 3/k → k^2 = 36 → k = ±6; check both in the third ratio.Test k = 6: k/12 = 6/12 = 1/2; 3/k = 3/6 = 1/2; (k − 3)/k = 3/6 = 1/2 → all equal.Test k = −6: k/12 = −1/2; 3/k = −1/2; (k − 3)/k = (−9)/(−6) = 3/2, not equal → reject.

Verification / Alternative check:
With k = 6, Equation 1 is 6x + 3y = 3 → divide by 6: x + (1/2)y = 1/2. Equation 2 is 12x + 6y = 6 → divide by 12: x + (1/2)y = 1/2. Same line → infinite solutions.

Why Other Options Are Wrong:
0, −6, 1, 3 fail the proportionality test across all three coefficients.

Common Pitfalls:
Matching only ax + by parts and forgetting to check constants; all three ratios must match.

For which value of k do the two linear equations kx + 3y = k − 3 and 12x + ky = k have infinitely many solutions?

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