Question

The value of \( k \), for which the linear equations \( 2x + 3y = 6 \) and \( 4x + 6y = 3k \) have at least one solution, is ________. (Answer in integer)

Hint:

For a system of linear equations to have at least one solution, the second equation must be a scalar multiple of the first equation.

Answer 1

Correct Answer: 4

We are given the system of linear equations: \[ 2x + 3y = 6 \quad {(Equation 1)} \] \[ 4x + 6y = 3k \quad {(Equation 2)}. \] To determine the value of \( k \) for which the system has at least one solution, we first observe that Equation 2 is just Equation 1 multiplied by 2: \[ 4x + 6y = 2(2x + 3y) = 2 \times 6 = 12. \] Thus, Equation 2 becomes: \[ 4x + 6y = 12. \] Now, for the system to have at least one solution, the right-hand sides of both equations must be consistent. Therefore, for Equation 2 to be consistent with Equation 1, we must have: \[ 3k = 12 \quad \Rightarrow \quad k = \frac{12}{3} = 4. \] Thus, the value of \( k \) is \( \boxed{4} \).