Question

Find the value of K for which the given system of equations has infinitely many solutions: \[ Kx + 3y = K-3 \quad \text{(1)} \] \[ 12x + Ky = K \quad \text{(2)} \]

Hint:

For infinitely many solutions, the system of equations must be proportional, and the ratio of corresponding coefficients must be equal.

Answer 1

Step 1: Conditions for infinitely many solutions.
For the system to have infinitely many solutions, the two equations must be proportional, meaning their corresponding coefficients must be proportional.
Step 2: Compare coefficients.
Comparing the coefficients of \( x \), \( y \), and the constant terms, we get the proportionality condition: \[ \frac{K}{12} = \frac{3}{K} = \frac{K-3}{K} \]
Step 3: Solve for K.
From the first ratio: \[ \frac{K}{12} = \frac{3}{K} \quad \Rightarrow \quad K^2 = 36 \quad \Rightarrow \quad K = 6 \, \text{or} \, K = -6 \]
Step 4: Check for consistency.
Substitute \( K = 6 \) into the second ratio: \[ \frac{K-3}{K} = \frac{6-3}{6} = \frac{3}{6} = \frac{1}{2} \] This is consistent with the other ratios. Therefore, the value of \( K \) that satisfies the condition is \( K = 6 \).