Question

Find the value of k for which the following system of equations has no solution.
$
  kx + 3y = 3 \\
  12x + ky = 6 \\
 $

Answer 1

Hint- Here, we will proceed by comparing the given pair of linear equations with any general pair of linear equations i.e., ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$. Then using the condition for having no solution i.e., $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$.

Complete Step-by-Step solution:
The given system of linear equations are $
  kx + 3y = 3 \\
   \Rightarrow kx + 3y - 3 = 0{\text{ }} \to {\text{(1)}} \\
 $ and $
  12x + ky = 6 \\
   \Rightarrow 12x + ky - 6 = 0{\text{ }} \to {\text{(2)}} \\
 $
As we know that for any pair of linear equations ${a_1}x + {b_1}y + {c_1} = 0{\text{ }} \to {\text{(3)}}$ and ${a_2}x + {b_2}y + {c_2} = 0{\text{ }} \to {\text{(4)}}$ to have no solution (inconsistent solution), the condition which must be satisfied is that the ratio of the coefficients of x should be equal to the ratio of the coefficients of y which further should not be equal to the ratio of the constant terms in the pair of linear equations.
The condition is $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}{\text{ }} \to (5{\text{)}}$
By comparing equations (1) and (3), we get
${a_1} = k,{b_1} = 3,{c_1} = - 3$
By comparing equations (2) and (4), we get
${a_2} = 12,{b_2} = k,{c_2} = - 6$
For the given pair of linear equations to have no solution, equation (5) must be satisfied
By equation (5), we can write
$
  \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}} \\
   \Rightarrow \dfrac{k}{{12}} = \dfrac{3}{k} \ne \dfrac{{ - 3}}{{ - 6}}{\text{ }} \to {\text{(6)}} \\
 $
By equation (6), we can write
\[
   \Rightarrow \dfrac{k}{{12}} = \dfrac{3}{k} \\
   \Rightarrow {k^2} = 3 \times 12 = 36 \\
   \Rightarrow k = \pm \sqrt {36} \\
   \Rightarrow k = \pm 6 \\
 \]
Now, putting k = 6 in equation (6), we have
$
   \Rightarrow \dfrac{6}{{12}} = \dfrac{3}{6} \ne \dfrac{{ - 3}}{{ - 6}} \\
   \Rightarrow \dfrac{1}{2} = \dfrac{1}{2} \ne \dfrac{1}{2} \\
 $
The obtained equation is not true so k = 6 is neglected
Now, putting k = -6 in equation (6), we have
$
   \Rightarrow \dfrac{{ - 6}}{{12}} = \dfrac{3}{{ - 6}} \ne \dfrac{{ - 3}}{{ - 6}} \\
   \Rightarrow \dfrac{{ - 1}}{2} = \dfrac{1}{{ - 2}} \ne \dfrac{1}{2} \\
   \Rightarrow - \dfrac{1}{2} = - \dfrac{1}{2} \ne \dfrac{1}{2} \\
 $
The above equation is always true so we can say k = -6
Therefore, the required value of k is -6.

Note- A pair of linear equations which are given by ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$ can also have unique solution (consistent solution) if the condition $\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}$ is satisfied. Also. For these pair of linear equations to have infinitely many solutions, the condition $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$ should always be satisfied.