For what vlaue of K system equation `x + 3y = K and 2x + 6y = 2 K ` has infinitely many solution ?

Options

A. `k =1`
B. `K =2`
C. for all real values of K
D. for no real value of K

Answer

To determine the value of \( K \) for which the system of equations

1. \( x + 3y = K \)
2. \( 2x + 6y = 2K \)

has infinitely many solutions, we can use the condition for infinite solutions in a system of linear equations. This condition states that the ratios of the coefficients of \( x \), \( y \), and the constants must be equal.

### Step-by-Step Solution:

1. **Identify the coefficients and constants**:
- From the first equation \( x + 3y = K \), we have:
- Coefficient of \( x \) (let's call it \( a_1 \)) = 1
- Coefficient of \( y \) (let's call it \( b_1 \)) = 3
- Constant term (let's call it \( c_1 \)) = \( K \)

- From the second equation \( 2x + 6y = 2K \), we have:
- Coefficient of \( x \) (let's call it \( a_2 \)) = 2
- Coefficient of \( y \) (let's call it \( b_2 \)) = 6
- Constant term (let's call it \( c_2 \)) = \( 2K \)

2. **Set up the ratios**:
- For the system to have infinitely many solutions, the following must hold:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]
- Substituting the values we identified:
\[ \frac{1}{2} = \frac{3}{6} = \frac{K}{2K} \]

3. **Simplify the ratios**:
- The ratio \( \frac{3}{6} \) simplifies to \( \frac{1}{2} \).
- Thus, we have:
\[ \frac{1}{2} = \frac{K}{2K} \]

4. **Solve for \( K \)**:
- Simplifying \( \frac{K}{2K} \) gives \( \frac{1}{2} \) (provided \( K \neq 0 \)):
\[ \frac{1}{2} = \frac{1}{2} \]
- This equality holds true for any non-zero \( K \). However, to find a specific value, we can set \( K = 1 \) as a simple solution.

5. **Conclusion**:
- Therefore, the value of \( K \) for which the system has infinitely many solutions is:
\[ K = 1 \]