Evaluate for what value of k the system of equations 2x - y = 5 and 6x + ky = 15 has infinitely many solutions.

Options

A. 8
B. -3
C. 3
D. 6

Answer

To find the value of \( k \) for which the system of equations has infinitely many solutions, we start with the two equations:

1. \( 2x - y = 5 \)
2. \( 6x + ky = 15 \)

### Step 1: Rewrite the equations in standard form
We can rewrite both equations in the standard form \( Ax + By + C = 0 \).

For the first equation:
\[ 2x - y - 5 = 0 \quad \Rightarrow \quad 2x - y + (-5) = 0 \]
Here, \( A_1 = 2 \), \( B_1 = -1 \), and \( C_1 = -5 \).

For the second equation:
\[ 6x + ky - 15 = 0 \quad \Rightarrow \quad 6x + ky + (-15) = 0 \]
Here, \( A_2 = 6 \), \( B_2 = k \), and \( C_2 = -15 \).

### Step 2: Apply the condition for infinitely many solutions
For the system to have infinitely many solutions, the following condition must hold:
\[ \frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} \]

Substituting the values we found:
\[ \frac{2}{6} = \frac{-1}{k} = \frac{-5}{-15} \]

### Step 3: Simplify the ratios
First, simplify \( \frac{2}{6} \) and \( \frac{-5}{-15} \):
\[ \frac{2}{6} = \frac{1}{3}, \quad \frac{-5}{-15} = \frac{1}{3} \]

Thus, we have:
\[ \frac{1}{3} = \frac{-1}{k} \]

### Step 4: Solve for \( k \)
Cross-multiplying gives us:
\[ 1 \cdot k = -1 \cdot 3 \quad \Rightarrow \quad k = -3 \]

### Conclusion
The value of \( k \) for which the system of equations has infinitely many solutions is:
\[ \boxed{-3} \]