The number of values of c, for which the system of equation `(c+1)x+8y=4c, cx + (c +3)y= 3c-1` has no solution is :

Options

A. infinite
B. 1
C. 2
D. 3

Explanation

To determine the number of values of \( c \) for which the system of equations

\[ (c+1)x + 8y = 4c \]
\[ cx + (c + 3)y = 3c - 1 \]

has no solution, we need to analyze the conditions under which a system of linear equations has no solutions. This occurs when the determinant of the coefficient matrix is zero, and the corresponding augmented matrix has a non-zero determinant.

### Step 1: Write the Coefficient Matrix and the Augmented Matrix

The coefficient matrix \( A \) and the augmented matrix \( [A|B] \) can be represented as follows:

\[ A = \begin{pmatrix} c + 1 & 8 \\ c & c + 3 \end{pmatrix}, \quad B = \begin{pmatrix} 4c \\ 3c - 1 \end{pmatrix} \]

### Step 2: Calculate the Determinant of the Coefficient Matrix

The determinant of matrix \( A \) is given by:

\[ \text{det}(A) = (c + 1)(c + 3) - (8)(c) \]

Expanding this, we get:

\[ \text{det}(A) = c^2 + 3c + c + 3 - 8c = c^2 - 4c + 3 \]

### Step 3: Set the Determinant Equal to Zero

To find the values of \( c \) for which the determinant is zero, we solve:

\[ c^2 - 4c + 3 = 0 \]

### Step 4: Factor the Quadratic Equation

Factoring the quadratic equation, we have:

\[ (c - 1)(c - 3) = 0 \]

Thus, the solutions are:

\[ c = 1 \quad \text{and} \quad c = 3 \]

### Step 5: Check the Condition for No Solutions

For the system to have no solutions, we need to check if the corresponding augmented matrix has a non-zero determinant when \( c = 1 \) and \( c = 3 \).

1. **For \( c = 1 \)**:
- The coefficient matrix becomes:
\[ A = \begin{pmatrix} 2 & 8 \\ 1 & 4 \end{pmatrix} \]
- The augmented matrix becomes:
\[ [A|B] = \begin{pmatrix} 2 & 8 & 4 \\ 1 & 4 & 2 \end{pmatrix} \]
- The determinant of the augmented matrix can be calculated, but since the determinant of \( A \) is zero, we need to check if the rows are proportional. They are not, hence no solution.

2. **For \( c = 3 \)**:
- The coefficient matrix becomes:
\[ A = \begin{pmatrix} 4 & 8 \\ 3 & 6 \end{pmatrix} \]
- The augmented matrix becomes:
\[ [A|B] = \begin{pmatrix} 4 & 8 & 12 \\ 3 & 6 & 8 \end{pmatrix} \]
- The determinant of the augmented matrix is also zero, and the rows are proportional, indicating no solution.

### Conclusion

Both values \( c = 1 \) and \( c = 3 \) result in the system having no solutions. Therefore, the number of values of \( c \) for which the system has no solution is:

\[ \boxed{2} \]