The simultaneous equations `3x + 5y = 7 and 6x + 10y = 18` have

Options

A. no solution
B. infinitely many solutions
C. unique solution
D. any finite number of solutions

Explanation

To determine the nature of the solutions for the simultaneous equations \(3x + 5y = 7\) and \(6x + 10y = 18\), we can analyze the equations step by step.

### Step 1: Write the equations in standard form
We can rewrite the equations in the form \(Ax + By + C = 0\).

1. The first equation \(3x + 5y = 7\) can be rewritten as:
\[ 3x + 5y - 7 = 0 \]

2. The second equation \(6x + 10y = 18\) can be rewritten as:
\[ 6x + 10y - 18 = 0 \]

### Step 2: Identify coefficients
From the equations, we can identify the coefficients:
- For the first equation \(3x + 5y - 7 = 0\):
- \(a_1 = 3\), \(b_1 = 5\), \(c_1 = -7\)

- For the second equation \(6x + 10y - 18 = 0\):
- \(a_2 = 6\), \(b_2 = 10\), \(c_2 = -18\)

### Step 3: Calculate the ratios of coefficients
We will compare the ratios of the coefficients \(a_1, b_1, c_1\) and \(a_2, b_2, c_2\):

1. Calculate \(\frac{a_1}{a_2}\):
\[ \frac{a_1}{a_2} = \frac{3}{6} = \frac{1}{2} \]

2. Calculate \(\frac{b_1}{b_2}\):
\[ \frac{b_1}{b_2} = \frac{5}{10} = \frac{1}{2} \]

3. Calculate \(\frac{c_1}{c_2}\):
\[ \frac{c_1}{c_2} = \frac{-7}{-18} = \frac{7}{18} \]

### Step 4: Analyze the ratios
Now we compare the ratios:
- We found that \(\frac{a_1}{a_2} = \frac{1}{2}\) and \(\frac{b_1}{b_2} = \frac{1}{2}\).
- However, \(\frac{c_1}{c_2} = \frac{7}{18}\).

According to the conditions for the solutions of simultaneous equations:
- If \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\), then the system has **no solution**.

### Conclusion
Since the conditions are satisfied, we conclude that the simultaneous equations \(3x + 5y = 7\) and \(6x + 10y = 18\) have **no solution**.