For what value of k, the system of equations x - 3y = 5 and 2x - 5y = k has at most one solution?

Answer

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

1. \( x - 3y = 5 \)
2. \( 2x - 5y = k \)

has at most one solution, we need to analyze the conditions for unique and no solutions in a system of linear equations.

### Step 1: Identify the coefficients
From the equations, we can identify the coefficients as follows:
- For the first equation \( x - 3y = 5 \):
- \( a_1 = 1 \)
- \( b_1 = -3 \)
- \( c_1 = 5 \)

- For the second equation \( 2x - 5y = k \):
- \( a_2 = 2 \)
- \( b_2 = -5 \)
- \( c_2 = k \)

### Step 2: Set up the ratios
We need to check the ratios of the coefficients:
- \( \frac{a_1}{a_2} = \frac{1}{2} \)
- \( \frac{b_1}{b_2} = \frac{-3}{-5} = \frac{3}{5} \)
- \( \frac{c_1}{c_2} = \frac{5}{k} \)

### Step 3: Conditions for unique and no solutions
1. **Unique Solution**: The system has a unique solution if \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \).
2. **No Solution**: The system has no solution if \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \) and \( \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \).

### Step 4: Analyze the ratios
From our calculations:
- \( \frac{a_1}{a_2} = \frac{1}{2} \)
- \( \frac{b_1}{b_2} = \frac{3}{5} \)

Since \( \frac{1}{2} \neq \frac{3}{5} \), we conclude that the system has a unique solution for all values of \( k \).

### Step 5: Conclusion
Thus, the system of equations will have at most one solution for **all real values of \( k \)**.