Write the number of solutions of the following pair of linear equations: `x+2y=3,\ \ \ \ 2x+4y=9`

Options

A. 0
B. 1
C. Infinite
D. none of these

Answer

To determine the number of solutions for the given pair of linear equations:

1. **Write down the equations:**
\[ x + 2y = 3 \quad \text{(1)} \]
\[ 2x + 4y = 9 \quad \text{(2)} \]

2. **Rewrite the equations in standard form:**
We can rewrite both equations in the standard form \(Ax + By + C = 0\):
\[ x + 2y - 3 = 0 \quad \text{(1)} \]
\[ 2x + 4y - 9 = 0 \quad \text{(2)} \]

3. **Identify coefficients:**
From the equations, we can identify:
- For equation (1): \(a_1 = 1\), \(b_1 = 2\), \(c_1 = -3\)
- For equation (2): \(a_2 = 2\), \(b_2 = 4\), \(c_2 = -9\)

4. **Calculate the ratios:**
Now, we will calculate the ratios:
\[ \frac{a_1}{a_2} = \frac{1}{2} \]
\[ \frac{b_1}{b_2} = \frac{2}{4} = \frac{1}{2} \]
\[ \frac{c_1}{c_2} = \frac{-3}{-9} = \frac{1}{3} \]

5. **Analyze the conditions:**
We have:
- \(\frac{a_1}{a_2} = \frac{1}{2}\)
- \(\frac{b_1}{b_2} = \frac{1}{2}\)
- \(\frac{c_1}{c_2} = \frac{1}{3}\)

We can see that:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \quad \text{but} \quad \frac{c_1}{c_2} \text{ is not equal to } \frac{a_1}{a_2} \text{ or } \frac{b_1}{b_2} \]

6. **Conclusion:**
Since \(\frac{a_1}{a_2} = \frac{b_1}{b_2}\) but \(\frac{c_1}{c_2} \neq \frac{a_1}{a_2}\), the equations are parallel and do not intersect. Therefore, there are no solutions.

Thus, the number of solutions for the given pair of linear equations is **zero**.

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