Question

If the system of linear equations
$\begin{array}{c}x-2y+kz=1\\ 2x+y+z=2\\ 3x-y-kz=3\end{array}$
has a solution $(x,y,z),z\ne 0$, then $(x,y)$ lies on the straight line whose equation is :

• A. $4x-3y-1=0$
• B. $3x-4y-1=0$
• C. $3x-4y-4=0$
• D. $4x-3y-4=0$

Answer 1

The correct option is D $4x-3y-4=0$

$x-2y+kz=1\cdots \left(1\right)$
$2x+y+z=2\cdots \left(2\right)$
$3x-y-kz=3\cdots \left(3\right)$

Adding $\left(1\right)$ and $\left(3\right)$, we get
$4x-3y=4$
$\Rightarrow 4x-3y-4=0$