Question & Answer
QUESTION

Find the values of K, for which the following pair of linear equations has infinitely many solutions.
$
  {\text{K}}x + 4y + \left( {\mu + 8} \right) = 0 \\
  4x + {\text{K}}y + 4 = 0 \\
 $

ANSWER Verified Verified
Hint: Here, we will proceed by using the condition for two linear equations which are ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$ to have an infinitely many solutions i.e., $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$ and then equating any pair of this equation.

Complete step-by-step answer:

Given, the linear equations are $
  {\text{K}}x + 4y + \left( {\mu + 8} \right) = 0{\text{ }} \to {\text{(1)}} \\
  4x + {\text{K}}y + 4 = 0{\text{ }} \to {\text{(2)}} \\
 $
For two linear equations i.e., ${a_1}x + {b_1}y + {c_1} = 0{\text{ }} \to {\text{(3)}}$ and ${a_2}x + {b_2}y + {c_2} = 0{\text{ }} \to {\text{(4)}}$ to have infinite number of solutions, the ratio of the coefficient of variable x must be equal to the ratio of the coefficients of variable y which further must be equal to the ratio of the constant terms.
i.e., $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}{\text{ }} \to {\text{(5)}}$
By comparing equation (1) with equation (3), we get
${a_1} = {\text{K,}}{b_1} = 4,{c_1} = \mu + 8$
By comparing equation (2) with equation (4), we get
${a_2} = 4{\text{,}}{b_2} = {\text{k}},{c_2} = 4$
By putting all the obtained values in the equation (5), we have
$\dfrac{{\text{K}}}{4} = \dfrac{4}{{\text{K}}} = \dfrac{{\mu + 8}}{4}{\text{ }} \to {\text{(6)}}$
$\dfrac{{\text{K}}}{4} = \dfrac{{\mu + 8}}{4}$
Taking $\dfrac{{\text{K}}}{4} = \dfrac{4}{{\text{K}}}$ from equation (6) and applying cross multiplication, we have
\[
   \Rightarrow {{\text{K}}^2} = 4 \times 4 = 16 \\
   \Rightarrow {\text{K}} = \pm \sqrt {16} \\
   \Rightarrow {\text{K}} = \pm 4 \\
 \]
Taking $\dfrac{4}{{\text{K}}} = \dfrac{{\mu + 8}}{4}$ from equation (6) and applying cross multiplication, we have
$
   \Rightarrow {\text{K}}\left( {\mu + 8} \right) = 4 \times 4 = 16 \\
   \Rightarrow \mu + 8 = \dfrac{{16}}{{\text{K}}}{\text{ }} \to {\text{(7)}} \\
 $
By putting K = 4 in equation (7), we get
\[
   \Rightarrow \mu + 8 = \dfrac{{16}}{{\text{4}}} = 4 \\
   \Rightarrow \mu = 4 - 8 \\
   \Rightarrow \mu = - 4 \\
 \]
By putting K = -4 in equation (7), we get
\[
   \Rightarrow \mu + 8 = \dfrac{{16}}{{ - {\text{4}}}} = - 4 \\
   \Rightarrow \mu = - 4 - 8 \\
   \Rightarrow \mu = - 12 \\
 \]
So, when K = 4, \[\mu = - 4\] and when K = -4, \[\mu = - 12\].

Therefore, the required values of K for which the given pair of linear equations has infinitely many solutions are \[{\text{K}} = \pm 4\].

Note: In this particular problem, in order to get the value of K it is necessary to consider $\dfrac{{\text{K}}}{4} = \dfrac{4}{{\text{K}}}$ whereas for the value of \[\mu \], we can also consider $\dfrac{{\text{K}}}{4} = \dfrac{{\mu + 8}}{4}$ because this will also give the same values of \[\mu \] i.e., when K = 4, \[\mu = - 4\] and when K = -4, \[\mu = - 12\]. If a pair of linear equations are given, there can also occur no solution or unique solution apart from infinitely many solutions.