For how many values of k. will the system of equations `(k + 1) x + 8y = 4k and kx + (k + 3) y = 3k - 1`, have an infinite number of solutions

Options

A. 1
B. 2
C. 3
D. None of these

Answer

To determine how many values of \( k \) will make the system of equations have an infinite number of solutions, we need to analyze the given equations:

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

### Step 1: Set up the condition for infinite solutions
For the system of equations to have an infinite number of solutions, the ratios of the coefficients must be equal. This means:

\[ \frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} \]

Where:
- \( A_1 = k + 1 \), \( A_2 = k \)
- \( B_1 = 8 \), \( B_2 = k + 3 \)
- \( C_1 = 4k \), \( C_2 = 3k - 1 \)

### Step 2: Set up the equations from the ratios
From the ratios, we can set up the following equations:

1. \( \frac{k + 1}{k} = \frac{8}{k + 3} \)
2. \( \frac{k + 1}{k} = \frac{4k}{3k - 1} \)

We can solve either of these equations to find \( k \). Let's solve the first equation.

### Step 3: Cross-multiply the first equation
Cross-multiplying gives us:

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

### Step 4: Expand and rearrange the equation
Expanding the left side:

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

This simplifies to:

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

Rearranging gives:

\[ k^2 + 4k - 8k + 3 = 0 \]

Which simplifies to:

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

### Step 5: Factor the quadratic equation
Now we factor the quadratic equation:

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

### Step 6: Solve for \( k \)
Setting each factor to zero gives us:

1. \( k - 3 = 0 \) → \( k = 3 \)
2. \( k - 1 = 0 \) → \( k = 1 \)

### Conclusion
Thus, there are **two values of \( k \)** (namely \( k = 1 \) and \( k = 3 \)) for which the system of equations will have an infinite number of solutions.