Solveeit Logo

Question

Question: Find the value of k or which of the following systems of equations be inconsistent. \(2x + ky + k...

Find the value of k or which of the following systems of equations be inconsistent.
2x+ky+k+2=0;kx+8y+3k=02x + ky + k + 2 = 0; kx + 8y + 3k = 0
A) For k=2k = 2 , the given system of equations will have no solution.
B) For k=4k = - 4 , the given system of equations will have no solution.
C) For k=6k = 6 , the given system of equations will have no solution.
D) For k=8k = - 8 , the given system of equations will have no solution.

Explanation

Solution

Here, we have to separate the value of a, b, and c from the given equation
For the equation to be inconsistent,
a1a2=b1b2c1c2\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=02x + ky + k + 2 = 0 and kx+8y+3k=0kx + 8y + 3k = 0
Consider the 1st{1^{st}} equation 2x+ky+k+2=02x + ky + k + 2 = 0
On rearranging, we get
ky=2xk2\Rightarrow ky = - 2x - k - 2
Now we can divide throughout with k. so, we get
y=2kx12k\Rightarrow y = \dfrac{{ - 2}}{k}x - 1 - \dfrac{2}{k}
Now it is of the form y=mx+cy = mx + c , where m is the slope of the line.
m1=2k\Rightarrow {m_1} = \dfrac{{ - 2}}{k}
Now consider the second equation, kx+8y+3k=0kx + 8y + 3k = 0
On rearranging, we get
8y=kx3k\Rightarrow 8y = - kx - 3k
Now we can divide throughout with k. so, we get
y=k8xk8\Rightarrow y = \dfrac{{ - k}}{8}x - \dfrac{k}{8}
Now it is of the form y=mx+cy = mx + c , where m is the slope of the line.
m2=k8\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.
m1=m2\Rightarrow {m_1} = {m_2}
On substituting the values, we get
2k=k8\Rightarrow \dfrac{{ - 2}}{k} = \dfrac{{ - k}}{8}
On cross multiplication, we get
k2=2×8\Rightarrow - {k^2} = - 2 \times 8
On cancelling the negative signs, we get
k2=16\Rightarrow {k^2} = 16
On taking the square root, we get
k=±16\Rightarrow k = \pm \sqrt {16}
Hence, we have
k=±4\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=4k = - 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=3x + y + z = 3
x+y+z=4x + y + z = 4
has no solution, as can be seen by subtracting the first equation from the second to obtain the impossible 0=10 = 1