Question

For what value of \( k \) equations \( kx + y = 1 \) and \( (k-1)x + 2y = 3 \) have no solution?

Hint:

Inconsistent Linear Equations: If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), then the system has no solution.

Answer 1

For no solution, the system of equations should be inconsistent, i.e., \[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \] From the given equations:
\[ \frac{k}{k-1} = \frac{1}{2} \neq \frac{1}{3} \] Solving:
\[ 2k = k-1 \] \[ 2k - k = -1 \] \[ k = -1 \] Thus, for \( k = -1 \), the system has no solution.
Correct Answer: \( k = -1 \)