Question

Find the values of k for which the system $
  2x + ky = 1 \\
  3x - 5y = 7 \\
 $ will have $\left( {\text{i}} \right)$ a unique solution, and $\left( {{\text{ii}}} \right)$ no solution. Is there a value of k for which the system has infinitely many solutions?

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 $
  2x + ky = 1 \\
   \Rightarrow 2x + ky - 1 = 0{\text{ }} \to {\text{(1)}} \\
 $ and $
  3x - 5y = 7 \\
   \Rightarrow 3x - 5y - 7 = 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 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
${a_1} = 2,{b_1} = k,{c_1} = - 1$
By comparing equations (2) and (4), we get
${a_2} = 3,{b_2} = - 5,{c_2} = - 7$
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{2}{3} \ne \dfrac{k}{{ - 5}} \\
   \Rightarrow k \ne \left( { - 5} \right) \times \left( {\dfrac{2}{3}} \right) \\
   \Rightarrow k \ne - \dfrac{{10}}{3} \\
 $
Therefore, for the given linear system of equations to have unique solution, the required value of k can be any number except $ - \dfrac{{10}}{3}$ i.e., $k \in \left( { - \infty , + \infty } \right) - \left\{ { - \dfrac{{10}}{3}} \right\}$.
$\left( {{\text{ii}}} \right)$ 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 $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}{\text{ }} \to (6{\text{)}}$
Here also, ${a_1} = 2,{b_1} = k,{c_1} = - 1$and ${a_2} = 3,{b_2} = - 5,{c_2} = - 7$
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{2}{3} = \dfrac{k}{{ - 5}} \ne \dfrac{{ - 1}}{{ - 7}}{\text{ }} \to {\text{(7)}} \\
 $
By equation (7), we can write
\[
   \Rightarrow \dfrac{2}{3} = \dfrac{k}{{ - 5}} \\
   \Rightarrow k = \left( { - 5} \right) \times \left( {\dfrac{2}{3}} \right) \\
   \Rightarrow k = - \dfrac{{10}}{3} \\
 \]
Now, putting k = $ - \dfrac{{10}}{3}$ in equation (6), we have
$
   \Rightarrow \dfrac{2}{3} = \dfrac{{\left( { - \dfrac{{10}}{3}} \right)}}{{ - 5}} \ne \dfrac{{ - 1}}{{ - 7}} \\
   \Rightarrow \dfrac{2}{3} = \dfrac{{10}}{{5 \times 3}} \ne \dfrac{1}{7} \\
   \Rightarrow \dfrac{2}{3} = \dfrac{2}{3} \ne \dfrac{1}{7} \\
 $
which is always true.
Therefore, for the given linear system of equations to have no solution, the required value of k is $ - \dfrac{{10}}{3}$

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.