First slide

System of linear equations

Question

If the system of linear equations $x - 2 y + k z = 1, 2 x + y + z = 2, 3 x - y - k z = 3$ has a solution $(x, y, x), z \neq o$ , then $(x, y)$ lies on the straight line whose equation is

Moderate

A

$3 x - 4 y - 4 = 0$
B

$3 x - 4 y - 1 = 0$
C

$4 x - 3 y - 4 = 0$
D

$4 x - 3 y - 1 = 0$

Solution

Given system of linear equations
$\begin{aligned} x - 2 y + k z = 1 - - - - (1) \\ 2 x + y + z = 2 - - - - (2) \end{aligned}$
$And 3 x - y - k z = 3 - - - - (3)$
$Has a solution (x, y, z), z \neq 0$
On adding Eqs. (i) and (iii), we get
$\begin{array}{l} x + 2 y + k z + 3 x - y - k z = 1 + 3 \\ 4 x - 3 y = 4 \\ \Rightarrow 4 x - 3 y - 4 = 0 \end{array}$
This is the required equation of the straight line in which point $(x, y)$ lies .

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