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. 1
B. 2
C. 3
D. infinite

Answer

To determine the number of values of \( k \) for which the system of equations has no solution, we need to analyze the given equations:

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

### Step 1: Identify coefficients
From the equations, we can identify the coefficients:
- 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 we identified:
\[ \frac{k + 1}{k} = \frac{8}{k + 3} \]

### Step 3: Cross-multiply to solve for \( k \)
Cross-multiplying gives:
\[ (k + 1)(k + 3) = 8k \]

Expanding the left side:
\[ k^2 + 3k + k + 3 = 8k \]
\[ k^2 + 4k + 3 = 8k \]

Rearranging the equation:
\[ k^2 + 4k + 3 - 8k = 0 \]
\[ k^2 - 4k + 3 = 0 \]

### Step 4: Factor the quadratic equation
We can factor the quadratic:
\[ (k - 1)(k - 3) = 0 \]

### Step 5: Find the roots
Setting each factor to zero gives us the potential values for \( k \):
\[ k - 1 = 0 \quad \Rightarrow \quad k = 1 \]
\[ k - 3 = 0 \quad \Rightarrow \quad k = 3 \]

### Step 6: Check the condition for no solution
Now we need to check if these values satisfy the condition for no solution:
1. For \( k = 1 \):
- \( \frac{b_1}{b_2} = \frac{8}{1 + 3} = \frac{8}{4} = 2 \)
- \( \frac{c_1}{c_2} = \frac{4 \cdot 1}{3 \cdot 1 - 1} = \frac{4}{2} = 2 \)
- Since \( \frac{b_1}{b_2} = \frac{c_1}{c_2} \), this does not satisfy the no solution condition.

2. For \( k = 3 \):
- \( \frac{b_1}{b_2} = \frac{8}{3 + 3} = \frac{8}{6} = \frac{4}{3} \)
- \( \frac{c_1}{c_2} = \frac{4 \cdot 3}{3 \cdot 3 - 1} = \frac{12}{8} = \frac{3}{2} \)
- Since \( \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), this satisfies the no solution condition.

### Conclusion
Thus, the only value of \( k \) for which the system of equations has no solution is \( k = 3 \). Therefore, the number of such values of \( k \) is:

**Answer: 1**