Question

For some constant real numbers p, k and a, consider the following system of linear equations in  x  and  y: 
px - 4y = 2 
3x + ky = a 
A necessary condition for the system to have no solution for  (x, y), is

• A. ap - 6 = 0
• B. \(kp + 12 \neq 0\)
• C. ap + 6 = 0
• D. \(2a + k \neq 0\)

Answer 1

The Correct Option is D

For the system of linear equations to have no solution, the lines represented by the equations must be parallel and not coincide. The condition for parallelism in a system of two linear equations $Ax + By = C$ and $Dx + Ey = F$ is that the ratio of the coefficients of $x$ and $y$ in both equations must be equal, i.e.,
\[ \frac{p}{-4} = \frac{3}{k} \]
This implies:
\[ p \cdot k = -12 \quad (1) \]
For no solution, the system should also not coincide, meaning the constant terms must not satisfy the same ratio. For this, we must have:
\[ \frac{2}{p} \ne \frac{a}{3} \]
Simplifying gives:
\[ 2a + k \ne 0 \quad (2) \]
Thus, the necessary condition for the system to have no solution is $2a + k \ne 0$, which corresponds to Option (4).