The value of c for which the pair of equations `cx - y = 2` and `6x - 2y =3` will have infinitely many solutions is

Options

A. 2
B. -3
C. -12
D. no value

Explanation

To find the value of \( c \) for which the pair of equations \( cx - y = 2 \) and \( 6x - 2y = 3 \) will have infinitely many solutions, we will use the condition for infinitely many solutions in a system of linear equations.

### Step-by-Step Solution:

1. **Write the equations in standard form:**
- The first equation is given as \( cx - y = 2 \). We can rewrite it as:
\[ cx - y - 2 = 0 \quad \text{(Equation 1)} \]
- The second equation is \( 6x - 2y = 3 \). We can rewrite it as:
\[ 6x - 2y - 3 = 0 \quad \text{(Equation 2)} \]

2. **Identify coefficients:**
- From Equation 1, we have:
- \( a_1 = c \)
- \( b_1 = -1 \)
- \( c_1 = -2 \)
- From Equation 2, we have:
- \( a_2 = 6 \)
- \( b_2 = -2 \)
- \( c_2 = -3 \)

3. **Set up the condition for infinitely many solutions:**
- The condition for infinitely many solutions is:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]
- Substituting the values we identified:
\[ \frac{c}{6} = \frac{-1}{-2} = \frac{-2}{-3} \]

4. **Solve the first equality:**
- From \( \frac{c}{6} = \frac{1}{2} \):
\[ c = 6 \cdot \frac{1}{2} = 3 \]

5. **Solve the second equality:**
- From \( \frac{c}{6} = \frac{2}{3} \):
\[ 3c = 2 \cdot 6 \]
\[ 3c = 12 \implies c = \frac{12}{3} = 4 \]

6. **Conclusion:**
- We found two values for \( c \): \( c = 3 \) and \( c = 4 \). However, for the equations to have infinitely many solutions, both conditions must hold true simultaneously, which is not possible. Therefore, there is no single value of \( c \) for which the equations have infinitely many solutions.

### Final Answer:
The value of \( c \) for which the pair of equations will have infinitely many solutions is **none**.