The number of values of k, for which the system of equations `(k""+""1)x""+""8y""=""4k` `k x""+""(k""+""3)y""=""3k-1` has no solution, is (1) 1 (2) 2 (3) 3 (4) infinite

Options

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

Explanation

To solve the problem, we need to determine the number of values of \( k \) for which the given system of equations has no solution. The equations are:

1. \( (k + 1)x + 8y = 4k \)
2. \( kx + (k + 3)y = 3k - 1 \)

### Step 1: Identify the coefficients
We can express the equations in the standard form \( ax + by = c \):

- For the first equation:
- \( a_1 = k + 1 \)
- \( b_1 = 8 \)
- \( c_1 = 4k \)

- For the second equation:
- \( a_2 = k \)
- \( b_2 = k + 3 \)
- \( c_2 = 3k - 1 \)

### Step 2: Apply the condition for no solution
The condition for the system of equations to have no solution is given by:

\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \]

Substituting the coefficients into the condition:

\[ \frac{k + 1}{k} = \frac{8}{k + 3} \neq \frac{4k}{3k - 1} \]

### Step 3: Solve the first part of the condition
We first solve the equality \( \frac{k + 1}{k} = \frac{8}{k + 3} \).

Cross-multiplying gives:

\[ (k + 1)(k + 3) = 8k \]

Expanding both sides:

\[ k^2 + 4k + 3 = 8k \]

Rearranging the equation:

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

### Step 4: Factor the quadratic equation
We can factor this quadratic:

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

Setting each factor to zero gives:

\[ k - 1 = 0 \quad \Rightarrow \quad k = 1 \]
\[ k - 3 = 0 \quad \Rightarrow \quad k = 3 \]

### Step 5: Check the second part of the condition
Now we need to check if these values satisfy the second part of the condition \( \frac{c_1}{c_2} \).

1. For \( k = 1 \):

\[ \frac{c_1}{c_2} = \frac{4(1)}{3(1) - 1} = \frac{4}{2} = 2 \]

2. For \( k = 3 \):

\[ \frac{c_1}{c_2} = \frac{4(3)}{3(3) - 1} = \frac{12}{8} = \frac{3}{2} \]

### Step 6: Compare the results
- For \( k = 1 \):
\[ \frac{k + 1}{k} = \frac{2}{1} = 2 \quad \text{and} \quad \frac{c_1}{c_2} = 2 \]
(This does not satisfy the condition \( \neq \))

- For \( k = 3 \):
\[ \frac{k + 1}{k} = \frac{4}{3} \quad \text{and} \quad \frac{c_1}{c_2} = \frac{3}{2} \]
(This satisfies the condition \( \neq \))

### Conclusion
The only value of \( k \) for which the system of equations has no solution is \( k = 3 \).

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