Evaluate for what value of c the system of linear equations `cx+3y=3, 12x+cy=6` has no solution.

Options

A. `-6`
B. 0
C. 6
D. 12

Answer

To determine the value of \( c \) for which the system of linear equations has no solution, we need to analyze the equations given:

1. The equations are:
\[ cx + 3y = 3 \quad \text{(1)} \]
\[ 12x + cy = 6 \quad \text{(2)} \]

2. For the system of equations to have no solution, the ratios of the coefficients of \( x \) and \( y \) must be equal, while the ratio of the constants must be different. This can be expressed as:
\[ \frac{A_1}{A_2} = \frac{B_1}{B_2} \neq \frac{C_1}{C_2} \]
where \( A_1 = c \), \( A_2 = 12 \), \( B_1 = 3 \), \( B_2 = c \), \( C_1 = 3 \), and \( C_2 = 6 \).

3. Setting up the equation:
\[ \frac{c}{12} = \frac{3}{c} \]

4. Cross-multiplying gives:
\[ c \cdot c = 3 \cdot 12 \]
\[ c^2 = 36 \]

5. Taking the square root of both sides, we find:
\[ c = 6 \quad \text{or} \quad c = -6 \]

6. Now we need to check the condition for no solution:
- We need to ensure that \( \frac{C_1}{C_2} \) is not equal to \( \frac{A_1}{A_2} \) or \( \frac{B_1}{B_2} \) for both values of \( c \).

7. For \( c = 6 \):
- \( \frac{A_1}{A_2} = \frac{6}{12} = \frac{1}{2} \)
- \( \frac{B_1}{B_2} = \frac{3}{6} = \frac{1}{2} \)
- \( \frac{C_1}{C_2} = \frac{3}{6} = \frac{1}{2} \) (equal, so this does not satisfy the condition)

8. For \( c = -6 \):
- \( \frac{A_1}{A_2} = \frac{-6}{12} = -\frac{1}{2} \)
- \( \frac{B_1}{B_2} = \frac{3}{-6} = -\frac{1}{2} \)
- \( \frac{C_1}{C_2} = \frac{3}{6} = \frac{1}{2} \) (not equal, so this satisfies the condition)

Thus, the value of \( c \) for which the system of equations has no solution is:
\[ \boxed{-6} \]