Question

The pair of linear equations $2kx + 5y = 7$, $6x - 5y = 11$ has a unique solution if
A. $k \ne - 3$
B. $k \ne 3$
C. $k \ne 5$
D. $k \ne - 5$

Answer 1

First we will first use that the system of linear equations ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$ will have a unique solution if $\dfrac{{{a_1}}}{{{b_1}}} = \dfrac{{{a_2}}}{{{b_2}}}$ and then we will find the value of ${a_1}$,${b_1}$ , ${c_1}$, ${a_2}$, ${b_2}$, and ${c_2}$ from the given system of equation. Then we will substitute the obtained values in the sufficient equation for unique solution.

Complete step by step answer:

We are given that the pair of linear equations
$2kx + 5y = 7{\text{ ......eq.(1)}}$
$6x - 5y = 11{\text{ ......eq(2)}}$
We know that the system of linear equations ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$ will have a unique solution if $\dfrac{{{a_1}}}{{{b_1}}} = \dfrac{{{a_2}}}{{{b_2}}}$.
Finding the value of ${a_1}$,${b_1}$ , ${c_1}$, ${a_2}$, ${b_2}$, and ${c_2}$ from the equation (1) and equation (2), we get
$\Rightarrow {a_1} = 2k$
$\Rightarrow {b_1} = 5$
$\Rightarrow {c_1} = - 7$
$\Rightarrow {a_2} = 6$
$\Rightarrow {b_2} = - 5$
$\Rightarrow {c_2} = - 11$
Substituting the above values in the sufficient equation for a unique solution, we get

$$\Rightarrow \dfrac{{2k}}{6} \ne \dfrac{5}{{ - 5}} \\\ \Rightarrow \dfrac{k}{3} \ne - 1 \\\$$

Multiplying the above equation by 3 on both sides, we get
$\Rightarrow k \ne - 3$

Hence, option A is correct.

Note: In solving this type of question, the key concept is to know that a system of linear equations ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$ will have a unique solution if $\dfrac{{{a_1}}}{{{b_1}}} = \dfrac{{{a_2}}}{{{b_2}}}$. This a simple problem, take care of calculations.