Find the value of k for which the system of equations `2x + 3y=7 and 8x + (k +4)y -28 =0` has infinitely many solution.

Answer

To find the value of \( k \) for which the system of equations

1. \( 2x + 3y = 7 \)
2. \( 8x + (k + 4)y - 28 = 0 \)

has infinitely many solutions, we need to determine the conditions under which the two equations represent the same line. This occurs when the ratios of the coefficients of \( x \), \( y \), and the constant terms are equal.

### Step 1: Rewrite the equations in standard form
The first equation is already in standard form:
\[ 2x + 3y - 7 = 0 \]

For the second equation, we rewrite it as:
\[ 8x + (k + 4)y - 28 = 0 \]

### Step 2: Identify coefficients
From the first equation \( 2x + 3y - 7 = 0 \):
- \( A_1 = 2 \)
- \( B_1 = 3 \)
- \( C_1 = -7 \)

From the second equation \( 8x + (k + 4)y - 28 = 0 \):
- \( A_2 = 8 \)
- \( B_2 = k + 4 \)
- \( C_2 = -28 \)

### Step 3: Set up the ratio condition for infinitely many solutions
For the two equations to have infinitely many solutions, the following condition must hold:
\[ \frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} \]

### Step 4: Calculate the ratios
1. Calculate \( \frac{A_1}{A_2} \):
\[ \frac{A_1}{A_2} = \frac{2}{8} = \frac{1}{4} \]

2. Calculate \( \frac{C_1}{C_2} \):
\[ \frac{C_1}{C_2} = \frac{-7}{-28} = \frac{1}{4} \]

3. Set \( \frac{B_1}{B_2} \) equal to \( \frac{1}{4} \):
\[ \frac{3}{k + 4} = \frac{1}{4} \]

### Step 5: Cross-multiply to solve for \( k \)
Cross-multiplying gives:
\[ 3 \cdot 4 = 1 \cdot (k + 4) \]
\[ 12 = k + 4 \]

### Step 6: Solve for \( k \)
Subtract 4 from both sides:
\[ k = 12 - 4 = 8 \]

### Conclusion
The value of \( k \) for which the system of equations has infinitely many solutions is:
\[ \boxed{8} \]