Find c if the system of equations `cx + 3y + (3-c)= 0, 12x + cy- c= 0` has infinitely many solutions?

Answer

To find the value of \( c \) such that the system of equations

\[ cx + 3y + (3 - c) = 0 \]
\[ 12x + cy - c = 0 \]

has infinitely many solutions, we will follow these steps:

### Step 1: Write the equations in standard form
The first equation can be rewritten as:
\[ cx + 3y + (3 - c) = 0 \implies cx + 3y = c - 3 \]

The second equation can be rewritten as:
\[ 12x + cy - c = 0 \implies 12x + cy = c \]

### Step 2: Identify coefficients
From the equations, we can identify the coefficients:
- For the first equation \( a_1 = c \), \( b_1 = 3 \), \( c_1 = 3 - c \)
- For the second equation \( a_2 = 12 \), \( b_2 = c \), \( c_2 = -c \)

### Step 3: Set up the condition for infinitely many solutions
For the system of equations to have infinitely many solutions, the following condition must hold:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]

Substituting the coefficients we identified:
\[ \frac{c}{12} = \frac{3}{c} = \frac{3 - c}{-c} \]

### Step 4: Solve the first ratio
First, we will solve the ratio \( \frac{c}{12} = \frac{3}{c} \):
Cross-multiplying gives:
\[ c^2 = 36 \implies c = 6 \text{ or } c = -6 \]

### Step 5: Solve the second ratio
Next, we will solve the ratio \( \frac{3}{c} = \frac{3 - c}{-c} \):
Cross-multiplying gives:
\[ 3(-c) = c(3 - c) \implies -3c = 3c - c^2 \]
Rearranging this gives:
\[ c^2 - 6c = 0 \implies c(c - 6) = 0 \]
Thus, \( c = 0 \) or \( c = 6 \).

### Step 6: Combine results
From the two sets of solutions, we have:
1. From the first ratio: \( c = 6 \) or \( c = -6 \)
2. From the second ratio: \( c = 0 \) or \( c = 6 \)

The common solution from both sets is:
\[ c = 6 \]

### Final Answer
Thus, the value of \( c \) for which the system of equations has infinitely many solutions is:
\[ \boxed{6} \]