Find the number of solutions for the pair of equations `x + 3y + 5 =0 and - 3x - 9y + 2 =0.`

Options

A. one
B. two
C. infinite
D. None

Answer

To find the number of solutions for the given pair of equations:

1. **Identify the equations**:
- First equation: \( x + 3y + 5 = 0 \)
- Second equation: \( -3x - 9y + 2 = 0 \)

2. **Rewrite the equations in standard form**:
- The first equation can be rewritten as:
\[ 1x + 3y + 5 = 0 \quad \text{(Here, } a_1 = 1, b_1 = 3, c_1 = 5\text{)} \]
- The second equation can be rewritten as:
\[ -3x - 9y + 2 = 0 \quad \text{(Here, } a_2 = -3, b_2 = -9, c_2 = 2\text{)} \]

3. **Calculate the ratios**:
- Calculate \( \frac{a_1}{a_2} \):
\[ \frac{a_1}{a_2} = \frac{1}{-3} = -\frac{1}{3} \]
- Calculate \( \frac{b_1}{b_2} \):
\[ \frac{b_1}{b_2} = \frac{3}{-9} = -\frac{1}{3} \]
- Calculate \( \frac{c_1}{c_2} \):
\[ \frac{c_1}{c_2} = \frac{5}{2} \]

4. **Analyze the ratios**:
- We have:
\[ \frac{a_1}{a_2} = -\frac{1}{3}, \quad \frac{b_1}{b_2} = -\frac{1}{3}, \quad \frac{c_1}{c_2} = \frac{5}{2} \]
- Here, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \) but \( \frac{c_1}{c_2} \) is not equal to the other two ratios.

5. **Conclusion**:
- According to the conditions for the number of solutions:
- If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \) but \( \frac{c_1}{c_2} \) is not equal to the other two, then the system of equations has **no solution**.

Thus, the number of solutions for the given pair of equations is **no solution**.