Question

Find the value of k for which the following system of equations has no solution.
$
  2x + ky = 11 \\
  5x - 7y = 5 \\
 $

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 $
  2x + ky = 11 \\
   \Rightarrow 2x + ky - 11 = 0{\text{ }} \to {\text{(1)}} \\
 $ and $
  5x - 7y = 5 \\
   \Rightarrow 5x - 7y - 5 = 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
${a_1} = 2,{b_1} = k,{c_1} = - 11$
By comparing equations (2) and (4), we get
${a_2} = 5,{b_2} = - 7,{c_2} = - 5$
For the given pair of linear equations to have no solution, equation (5) must be satisfied
Considering $\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$
 $
  \dfrac{2}{5} \ne \dfrac{{ - 11}}{{ - 5}} \\
   \Rightarrow \dfrac{2}{5} \ne \dfrac{{11}}{5} \\
 $
which is always true.
By equation (5), we can write
$
  \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \\
   \Rightarrow \dfrac{2}{5} = \dfrac{k}{{ - 7}} \\
   \Rightarrow k = \dfrac{2}{5}\left( { - 7} \right) \\
   \Rightarrow k = \dfrac{{ - 14}}{5} \\
 $
Therefore, the required value of k is $\dfrac{{ - 14}}{5}$.

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.