The system of equations `x + 2y = 13` and ` 3x + 6y = 9` has :

Options

A. Unique solution.
B. No solution
C. Infinitely many solutions.
D. Finite number of solutions.

Answer

To determine the nature of the solutions for the given system of equations \(x + 2y = 13\) and \(3x + 6y = 9\), we can follow these steps:

### Step 1: Write the equations in standard form
The equations are already in standard form:
1. \(x + 2y - 13 = 0\) (Equation 1)
2. \(3x + 6y - 9 = 0\) (Equation 2)

### Step 2: Identify coefficients
From the equations, we can identify the coefficients:
- For Equation 1:
- \(a_1 = 1\)
- \(b_1 = 2\)
- \(c_1 = -13\)

- For Equation 2:
- \(a_2 = 3\)
- \(b_2 = 6\)
- \(c_2 = -9\)

### Step 3: Calculate the ratios of the coefficients
Now we calculate the ratios:
1. \( \frac{a_1}{a_2} = \frac{1}{3} \)
2. \( \frac{b_1}{b_2} = \frac{2}{6} = \frac{1}{3} \)
3. \( \frac{c_1}{c_2} = \frac{-13}{-9} = \frac{13}{9} \)

### Step 4: Analyze the ratios
Now we compare the ratios:
- \( \frac{a_1}{a_2} = \frac{1}{3} \)
- \( \frac{b_1}{b_2} = \frac{1}{3} \)
- \( \frac{c_1}{c_2} = \frac{13}{9} \)

### Step 5: Determine the nature of the solutions
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \) but \( \frac{c_1}{c_2} \) is not equal to these ratios, we fall into the second case of the conditions for the system of equations:
- **Case**: \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \) but \( \frac{c_1}{c_2} \neq \frac{a_1}{a_2} \)
- **Conclusion**: This indicates that the system of equations has **no solution**.

### Final Answer:
The system of equations \(x + 2y = 13\) and \(3x + 6y = 9\) has **no solution**.

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