Question

The value of \[c\] for which the pair of equations \[cx + y = 2\] and \[6x + 2y = 3\] will have infinitely many solutions is
A. \[ - 3\]
B. \[3\]
C. \[12\]
D. None of these

Answer 1

Hint: First of all, rewrite the given pair of equations such that their RHS are equal to zero. As these pairs of equations have infinitely many solutions, use the condition for having infinitely many solutions to find the value of \[c\]. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer:

The given lines are \[cx + y = 2\] and \[6x + 2y = 3\] which can be rewrite as \[cx + y - 2 = 0\] & \[6x + 2y - 3 = 0\].
We know that the condition for two lines \[{a_1}x + {b_1}y + {c_1} = 0{\text{ and }}{a_2}x + {b_2}y + {c_2} = 0\] to have infinitely many solutions is \[\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}\]
Let \[{a_1}x + {b_1}y + {c_1} = cx + y - 2 = 0\]
And \[{a_2}x + {b_2}y + {c_2} = 6x + 2y - 3 = 0\]
So, we have \[{a_1} = c,{b_1} = 1,{c_1} = - 2{\text{ and }}{a_2} = 6,{b_2} = 2,{c_2} = - 3\]
As the given pair of equations have infinitely solutions, we have
\[\dfrac{c}{6} = \dfrac{1}{2} = \dfrac{{ - 2}}{{ - 3}}\]
Now taking \[\dfrac{c}{6} = \dfrac{1}{2}\], we get
\[
  \dfrac{c}{6} = \dfrac{1}{2} \\
  2c = 6 \\
  \therefore c = \dfrac{6}{2} = 3 \\
\]
And taking \[\dfrac{c}{6} = \dfrac{{ - 2}}{{ - 3}}\], we get
\[
  \dfrac{c}{6} = \dfrac{{ - 2}}{{ - 3}} \\
   - 3c = - 2 \times 6 \\
   - 3c = - 12 \\
  \therefore c = \dfrac{{ - 12}}{{ - 3}} = 4 \\
\]
Here, \[c\] has different values. If we put the values of \[c\] the condition for infinitely many solutions does not satisfy.
Thus, the correct option is D. None of these.

Note: The system of equations has infinitely many solutions when the lines are coincident, and they have the same y-intercept. The condition for two lines \[{a_1}x + {b_1}y + {c_1} = 0{\text{ and }}{a_2}x + {b_2}y + {c_2} = 0\] to have infinitely many solutions is \[\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}\].