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

Hint: 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.