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

Answer: (D)

$2a + k \neq 0$

Answer 2

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).