Question

Prove that there is a value of c($ \ne $0) for which the system $
  6x + 3y = c - 3 \\
  12x + cy = c \\
 $ 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 infinitely many solutions i.e., $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$.

Complete Step-by-Step solution:
The given system of linear equations is $
  6x + 3y = c - 3 \\
   \Rightarrow 6x + 3y - \left( {c - 3} \right) = 0 \to {\text{(1)}} \\
 $ and $
  12x + cy = c \\
   \Rightarrow 12x + cy - c = 0{\text{ }} \to {\text{(2)}} \\
 $
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 infinitely many solutions, 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 be equal to the ratio of the constant terms in the pair of linear equations.
The condition is $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}{\text{ }} \to (5{\text{)}}$
By comparing equations (1) and (3), we get
${a_1} = 6,{b_1} = 3,{c_1} = - \left( {c - 3} \right)$
By comparing equations (2) and (4), we get
${a_2} = 12,{b_2} = c,{c_2} = - c$
For the given pair of linear equations to have inconsistent solution, equation (5) must be satisfied
By equation (5), we can write
\[
  \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}} \\
   \Rightarrow \dfrac{6}{{12}} = \dfrac{3}{c} = \dfrac{{ - \left( {c - 3} \right)}}{{ - c}} \\
   \Rightarrow \dfrac{1}{2} = \dfrac{3}{c} = \dfrac{{c - 3}}{c}{\text{ }} \to {\text{(6)}} \\
 \]
By equation (6), we can write
\[
   \Rightarrow \dfrac{1}{2} = \dfrac{3}{c} \\
   \Rightarrow c = 3 \times 2 = 6 \\
 \]
By putting c = 6 in equation (6), we get
\[
   \Rightarrow \dfrac{1}{2} = \dfrac{3}{6} = \dfrac{{6 - 3}}{6} \\
   \Rightarrow \dfrac{1}{2} = \dfrac{1}{2} = \dfrac{3}{6} \\
   \Rightarrow \dfrac{1}{2} = \dfrac{1}{2} = \dfrac{1}{2} \\
 \]
which is always true.
Therefore, the required value of c for which the given system of linear equations has infinitely many solutions is 6.

Note- Any general 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 unique solution (consistent solution) if the condition $\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}$ is satisfied. Also, for these pair of linear equations to have no solution, the condition $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$ should always be satisfied.