Question

Find the value of k for which the following system $
  kx + 2y = 5 \\
  3x + y = 1 \\
 $ has
$\left( {\text{i}} \right)$ a unique solution and $\left( {{\text{ii}}} \right)$ no solution.

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 unique solution i.e., $\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}$and for 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 + 2y = 5 \\
   \Rightarrow kx + 2y - 5 = 0{\text{ }} \to {\text{(1)}} \\
 $ and $
  3x + y = 1 \\
   \Rightarrow 3x + y - 1 = 0{\text{ }} \to {\text{(2)}} \\
 $
$\left( {\text{i}} \right)$ 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 a unique solution (consistent solution), the condition which must be satisfied is that the ratio of the coefficients of x should not be equal to the ratio of the coefficients of y in the pair of linear equations.
The condition is $\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}{\text{ }} \to (5{\text{)}}$
By comparing equations (1) and (3), we get
${a_1} = k,{b_1} = 2,{c_1} = - 5$
By comparing equations (2) and (4), we get
${a_2} = 3,{b_2} = 1,{c_2} = - 1$
For the given pair of linear equations to have unique solution, equation (5) must be satisfied
By equation (5), we can write
$
  \dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}{\text{ }} \\
   \Rightarrow \dfrac{k}{3} \ne \dfrac{2}{1} \\
   \Rightarrow k \ne 2 \times 3 \\
   \Rightarrow k \ne 6 \\
 $
Therefore, for the given linear system of equations to have unique solution, the required value of k can be any number except 6 i.e., $k \in \left( { - \infty , + \infty } \right) - \left\{ 6 \right\}$.
$\left( {{\text{ii}}} \right)$ Also 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 \[\]
Here also, ${a_1} = k,{b_1} = 2,{c_1} = - 5$ and ${a_2} = 3,{b_2} = 1,{c_2} = - 1$
For the given pair of linear equations to have no solution, equation (6) must be satisfied
By equation (6), we can write
$
  \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}} \\
   \Rightarrow \dfrac{k}{3} = \dfrac{2}{1} \ne \dfrac{{ - 5}}{{ - 1}}{\text{ }} \to {\text{(7)}} \\
 $
By equation (7), we can write
\[
   \Rightarrow \dfrac{k}{3} = \dfrac{2}{1} \\
   \Rightarrow k = 3 \times \left( {\dfrac{2}{1}} \right) \\
   \Rightarrow k = 6 \\
 \]
Now, putting k = 6 in equation (6), we have
$
   \Rightarrow \dfrac{6}{3} = \dfrac{2}{1} \ne \dfrac{{ - 5}}{{ - 1}} \\
   \Rightarrow 2 = 2 \ne 5 \\
 $
which is always true.
Therefore, for the given linear system of equations to have no solution, 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 infinitely many solutions, the condition $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$ should always be satisfied. In this particular problem, the condition doesn’t depend on the constant terms for the pair of linear equations to have unique solutions.