The pair of equations `x + 2y + 5 = 0` and ` -3x - 6y + 1 = 0` has

Options

A. a unique solution
B. exactly two solutions
C. infinitely many solutions
D. no solution

Explanation

To determine the nature of the pair of equations given by \( x + 2y + 5 = 0 \) and \( -3x - 6y + 1 = 0 \), we will analyze the coefficients of the equations using the criteria for the existence of solutions.

### Step-by-step Solution:

1. **Rewrite the equations in standard form**:
- The first equation can be rewritten as:
\[ x + 2y + 5 = 0 \quad \Rightarrow \quad x + 2y = -5 \]
- The second equation can be rewritten as:
\[ -3x - 6y + 1 = 0 \quad \Rightarrow \quad -3x - 6y = -1 \quad \Rightarrow \quad 3x + 6y = 1 \]

2. **Identify coefficients**:
- For the first equation \( a_1 = 1, b_1 = 2, c_1 = -5 \)
- For the second equation \( a_2 = 3, b_2 = 6, c_2 = 1 \)

3. **Calculate the ratios**:
- Calculate \( \frac{a_1}{a_2} \):
\[ \frac{a_1}{a_2} = \frac{1}{3} \]
- Calculate \( \frac{b_1}{b_2} \):
\[ \frac{b_1}{b_2} = \frac{2}{6} = \frac{1}{3} \]
- Calculate \( \frac{c_1}{c_2} \):
\[ \frac{c_1}{c_2} = \frac{-5}{1} = -5 \]

4. **Analyze the ratios**:
- We observe that:
\[ \frac{a_1}{a_2} = \frac{1}{3}, \quad \frac{b_1}{b_2} = \frac{1}{3}, \quad \frac{c_1}{c_2} = -5 \]
- Here, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \) but \( \frac{c_1}{c_2} \) is not equal to these ratios.

5. **Conclusion**:
- According to the criteria for the solution of linear equations:
- If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \) but \( \frac{c_1}{c_2} \neq \frac{a_1}{a_2} \), then the system of equations has **no solution**.
- Therefore, the pair of equations given above has **no solution**.

### Final Answer:
The pair of equations \( x + 2y + 5 = 0 \) and \( -3x - 6y + 1 = 0 \) has **no solution**.