For what value of k, the system of equations x - 3y = 5 and 2x - ky = 10 has at most one solution?

Answer

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

1. \( x - 3y = 5 \)
2. \( 2x - ky = 10 \)

has at most one solution, we need to analyze the conditions for unique and no solutions.

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

- For the second equation \( 2x - ky = 10 \):
- \( a_2 = 2 \)
- \( b_2 = -k \)
- \( c_2 = -10 \)

### Step 2: Set up the ratios
For the system of equations to have at most one solution, we need to check the ratios of the coefficients:
- The condition for a unique solution is:
\[ \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \]

- The condition for no solution is:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \]

### Step 3: Calculate the ratios
Calculating the ratios:
- \( \frac{a_1}{a_2} = \frac{1}{2} \)
- \( \frac{b_1}{b_2} = \frac{-3}{-k} = \frac{3}{k} \)
- \( \frac{c_1}{c_2} = \frac{-5}{-10} = \frac{1}{2} \)

### Step 4: Analyze the unique solution condition
For a unique solution:
\[ \frac{1}{2} \neq \frac{3}{k} \]
Cross-multiplying gives:
\[ 1 \cdot k \neq 2 \cdot 3 \implies k \neq 6 \]

### Step 5: Analyze the no solution condition
For no solution, we need:
\[ \frac{1}{2} = \frac{3}{k} \quad \text{and} \quad \frac{1}{2} \neq \frac{1}{2} \]
The second part is always false, hence no solution condition does not apply here.

### Conclusion
Thus, the system of equations will have at most one solution for all real values of \( k \) except \( k = 6 \).

**Final Answer:** \( k \neq 6 \)

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