Question

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

Hint:

For parallel lines, the ratios of coefficients of \( x \) and \( y \) must be equal, but not the constant terms.

Answer 1

For a system of equations to have no solution, the lines must be parallel, which occurs when:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}. \] Comparing with \( ax + by = c \):
\[ \frac{k}{k+1} = \frac{1}{2} \neq \frac{1}{3}. \] Cross multiplying:
\[ 2k = k+1. \] \[ 2k - k = 1. \] \[ k = 1. \] Thus, for \( k = 1 \), the system has no solution.