Question

Find the value of k or which of the following systems of equations be inconsistent.
 $2x + ky + k + 2 = 0; kx + 8y + 3k = 0$
A) For \[k = 2\] , the given system of equations will have no solution.
B) For \[k = - 4\] , the given system of equations will have no solution.
C) For \[k = 6\] , the given system of equations will have no solution.
D) For \[k = - 8\] , the given system of equations will have no solution.

Answer 1

Hint:
Here, we have to separate the value of a, b, and c from the given equation
For the equation to be inconsistent,
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$
We have to put the value of a, b and in the above equation to find the value of k.

Complete step by step solution:
We are given the equations \[2x + ky + k + 2 = 0\] and $kx + 8y + 3k = 0$
Consider the \[{1^{st}}\] equation \[2x + ky + k + 2 = 0\]
On rearranging, we get
\[ \Rightarrow ky = - 2x - k - 2\]
Now we can divide throughout with k. so, we get
\[ \Rightarrow y = \dfrac{{ - 2}}{k}x - 1 - \dfrac{2}{k}\]
Now it is of the form $y = mx + c$ , where m is the slope of the line.
$ \Rightarrow {m_1} = \dfrac{{ - 2}}{k}$
Now consider the second equation, $kx + 8y + 3k = 0$
On rearranging, we get
\[ \Rightarrow 8y = - kx - 3k\]
Now we can divide throughout with k. so, we get
\[ \Rightarrow y = \dfrac{{ - k}}{8}x - \dfrac{k}{8}\]
Now it is of the form $y = mx + c$ , where m is the slope of the line.
$ \Rightarrow {m_2} = \dfrac{{ - k}}{8}$
It is given that the system of equations is inconsistent. As the solution of 2 equations is the point of intersection of the lines represented them, these lines will be parallel. So, we can equate the slopes.
$ \Rightarrow {m_1} = {m_2}$
On substituting the values, we get
$ \Rightarrow \dfrac{{ - 2}}{k} = \dfrac{{ - k}}{8}$
On cross multiplication, we get
 $ \Rightarrow - {k^2} = - 2 \times 8$
On cancelling the negative signs, we get
$ \Rightarrow {k^2} = 16$
On taking the square root, we get
$ \Rightarrow k = \pm \sqrt {16} $
Hence, we have
$ \Rightarrow k = \pm 4$
So, k can be either 4 or -4 for the equation to be inconsistent.

Therefore, the correct answer is option B which is, For, $k = - 4$, the given system of equations will have no solution.

Note:
Inconsistency: A linear or nonlinear equation is called inconsistent if there is no set of values for the unknown that satisfies all of the equations.
For example,
 $x + y + z = 3$
 $x + y + z = 4$
has no solution, as can be seen by subtracting the first equation from the second to obtain the impossible \[0 = 1\]