Question

For what value of \( k \) will the system of equations \( 2x + 3y = 12 \) and \( 4x + ky = 24 \) have no solution?

• A. 3
• B. 6
• C. 9
• D. 12
• E. 15

Hint:

For a system of linear equations to have no solution, the lines represented by the equations must be parallel. This happens when the ratios of the coefficients of \( x \) and \( y \) are equal but the constants are not.

Answer 1

The Correct Option is B

For the system of equations to have no solution, the two lines must be parallel. This occurs when the coefficients of \( x \) and \( y \) in both equations are proportional. The first equation is: \[ 2x + 3y = 12. \] The second equation is: \[ 4x + ky = 24. \] To find the condition for no solution, we compare the ratios of the coefficients of \( x \) and \( y \) in both equations. The coefficients of \( x \) in the two equations are \( 2 \) and \( 4 \), and the coefficients of \( y \) are \( 3 \) and \( k \). For the lines to be parallel, the ratios of the coefficients of \( x \) and \( y \) must be equal. Therefore, we set up the equation: \[ \frac{2}{4} = \frac{3}{k}. \] Simplifying the left-hand side: \[ \frac{1}{2} = \frac{3}{k}. \] Now, cross-multiply to solve for \( k \): \[ 1 \times k = 2 \times 3
\Rightarrow
k = 6. \] Thus, the value of \( k \) that makes the two lines parallel, and thus ensures no solution for the system of equations, is: \[ k = 6. \]