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 find 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 solution. This occurs when the lines represented by the equations are parallel, which means their slopes are equal but their intercepts are different.

### Step 1: Write the equations in slope-intercept form

We can rewrite both equations in the form \( y = mx + b \) to find their slopes.

1. From the first equation:
\[ 8y = 4c - (c + 1)x \]
\[ y = \frac{4c - (c + 1)x}{8} \]
\[ y = -\frac{(c + 1)}{8}x + \frac{4c}{8} \]
The slope \( m_1 = -\frac{(c + 1)}{8} \).

2. From the second equation:
\[ (c + 3)y = 3c - cx \]
\[ y = \frac{3c - cx}{c + 3} \]
\[ y = -\frac{c}{c + 3}x + \frac{3c}{c + 3} \]
The slope \( m_2 = -\frac{c}{c + 3} \).

### Step 2: Set the slopes equal for parallel lines

For the lines to be parallel, we set the slopes equal:
\[ -\frac{(c + 1)}{8} = -\frac{c}{c + 3} \]

### Step 3: Cross-multiply to eliminate the fractions

Cross-multiplying gives:
\[ (c + 1)(c + 3) = 8c \]

### Step 4: Expand and rearrange the equation

Expanding the left side:
\[ c^2 + 3c + c + 3 = 8c \]
\[ c^2 + 4c + 3 = 8c \]
Rearranging gives:
\[ c^2 - 4c + 3 = 0 \]

### Step 5: Factor the quadratic equation

Factoring the quadratic:
\[ (c - 1)(c - 3) = 0 \]

### Step 6: Solve for \( c \)

Setting each factor to zero gives:
\[ c - 1 = 0 \quad \Rightarrow \quad c = 1 \]
\[ c - 3 = 0 \quad \Rightarrow \quad c = 3 \]

### Step 7: Check for no solution condition

Now we need to check if these values of \( c \) lead to no solution. For no solution, the intercepts must be different.

1. For \( c = 1 \):
- The first equation becomes \( 2x + 8y = 4 \) or \( y = -\frac{1}{4}x + \frac{1}{2} \).
- The second equation becomes \( x + 4y = 2 \) or \( y = -\frac{1}{4}x + \frac{1}{2} \).
- Both equations are the same, hence they have infinitely many solutions.

2. For \( c = 3 \):
- The first equation becomes \( 4x + 8y = 12 \) or \( y = -\frac{1}{2}x + \frac{3}{2} \).
- The second equation becomes \( 3x + 6y = 8 \) or \( y = -\frac{1}{2}x + \frac{4}{3} \).
- The slopes are the same, but the intercepts are different, hence no solution.

### Conclusion

The only value of \( c \) for which the system has no solution is \( c = 3 \).

Thus, the number of values of \( c \) for which the system of equations has no solution is **1**.

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