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

Step 1: Write down the system of equations.
\[ \begin{aligned} 2x + 3y &= 6 \quad \text{(Equation 1)} \\ 4x + 6y &= 3k \quad \text{(Equation 2)} \end{aligned} \] Step 2: Analyze the relationship between the equations.
Notice that Equation 2 is exactly twice Equation 1 on the left-hand side: \[ 4x + 6y = 2(2x + 3y) \] So the left-hand side of Equation 2 is simply: \[ 4x + 6y = 2 \cdot 6 = 12 \] Step 3: Determine the condition for consistency.
For the two equations to represent the same line (and thus have infinitely many solutions or at least one), the right-hand side of Equation 2 must also be 12: \[ 3k = 12 \quad \Rightarrow \quad k = \frac{12}{3} = 4 \] Final Answer:
The system has at least one solution when \( \boxed{k = 4} \).