8. The value of \(c\) for which the pair of equations \(cx - y = 2\) and \(6x - 2y = 3\) will have infinitely many solutions is
3
–3
–12
no value
Step 1: Write equations in standard form.
First equation: \(cx - y = 2\) ⇒ \(cx - y - 2 = 0\).
So, \(a_1 = c, \; b_1 = -1, \; c_1 = -2\).
Second equation: \(6x - 2y = 3\) ⇒ \(6x - 2y - 3 = 0\).
So, \(a_2 = 6, \; b_2 = -2, \; c_2 = -3\).
Step 2: Condition for infinitely many solutions.
We need
\(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}.\)
Step 3: Compare ratios.
\(\dfrac{b_1}{b_2} = \dfrac{-1}{-2} = \dfrac{1}{2}\).
\(\dfrac{c_1}{c_2} = \dfrac{-2}{-3} = \dfrac{2}{3}\).
Step 4: Observe.
Since \(\dfrac{1}{2} \ne \dfrac{2}{3}\), the second and third ratios are not equal.
This inequality is true regardless of the value of \(c\). Hence, it is impossible to satisfy the condition for infinitely many solutions.
Conclusion. There is no value of \(c\) that makes the system have infinitely many solutions.