QUESTION

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

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.

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.