Evaluate for what value of c for which 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 find the value of \( c \) for which the system of linear equations

\[ cx + 3y = 3 \]
\[ 12x + cy = 6 \]

has no solution, we can follow these steps:

### Step 1: Rewrite the equations in standard form
We can rewrite both equations in the standard form \( Ax + By + C = 0 \).

1. For the first equation, \( cx + 3y = 3 \):
\[ cx + 3y - 3 = 0 \]
Here, \( A_1 = c \), \( B_1 = 3 \), and \( C_1 = -3 \).

2. For the second equation, \( 12x + cy = 6 \):
\[ 12x + cy - 6 = 0 \]
Here, \( A_2 = 12 \), \( B_2 = c \), and \( C_2 = -6 \).

### Step 2: Apply the condition for no solution
For the system of equations to have no solution, the following condition must hold:
\[ \frac{A_1}{A_2} = \frac{B_1}{B_2} \quad \text{and} \quad \frac{A_1}{A_2} \neq \frac{C_1}{C_2} \]

Substituting the values we have:
\[ \frac{c}{12} = \frac{3}{c} \]

### Step 3: Cross-multiply to solve for \( c \)
Cross-multiplying gives:
\[ c^2 = 36 \]

### Step 4: Solve for \( c \)
Taking the square root of both sides:
\[ c = 6 \quad \text{or} \quad c = -6 \]

### Step 5: Check the condition for no solution
We need to check if either value of \( c \) satisfies the condition \( \frac{A_1}{A_2} \neq \frac{C_1}{C_2} \).

1. For \( c = 6 \):
\[ \frac{A_1}{A_2} = \frac{6}{12} = \frac{1}{2} \]
\[ \frac{C_1}{C_2} = \frac{-3}{-6} = \frac{1}{2} \]
Since these are equal, \( c = 6 \) does not satisfy the condition.

2. For \( c = -6 \):
\[ \frac{A_1}{A_2} = \frac{-6}{12} = -\frac{1}{2} \]
\[ \frac{C_1}{C_2} = \frac{-3}{-6} = \frac{1}{2} \]
Since these are not equal, \( c = -6 \) satisfies the condition.

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