For what value of m, does the system of equations have a unique solution?
`x+ 2y= 3`
`5x+ my + 7= 0`

Options

A. `m ne 2`
B. `m ne 10`
C. `m ne 5`
D. `m ne - 10`

Answer

To find the value of \( m \) for which the system of equations has a unique solution, we need to analyze the given equations:

1. The first equation is:
\[ x + 2y = 3 \]

2. The second equation can be rewritten as:
\[ 5x + my + 7 = 0 \implies 5x + my = -7 \]

Now, we can express the system of equations in the standard form:
- \( a_1x + b_1y + c_1 = 0 \)
- \( a_2x + b_2y + c_2 = 0 \)

From the first equation, we can rewrite it as:
\[ x + 2y - 3 = 0 \implies a_1 = 1, b_1 = 2, c_1 = -3 \]

From the second equation, we have:
\[ 5x + my + 7 = 0 \implies a_2 = 5, b_2 = m, c_2 = 7 \]

For the system of equations to have a unique solution, the ratio of the coefficients must not be equal:
\[ \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \]

Calculating the ratios:
1. \( \frac{a_1}{a_2} = \frac{1}{5} \)
2. \( \frac{b_1}{b_2} = \frac{2}{m} \)
3. \( \frac{c_1}{c_2} = \frac{-3}{7} \)

Setting up the inequalities based on the conditions for a unique solution:
1. \( \frac{1}{5} \neq \frac{2}{m} \)
2. \( \frac{2}{m} \neq \frac{-3}{7} \)

### Step 1: Solve \( \frac{1}{5} \neq \frac{2}{m} \)
Cross-multiplying gives:
\[ 1 \cdot m \neq 2 \cdot 5 \implies m \neq 10 \]

### Step 2: Solve \( \frac{2}{m} \neq \frac{-3}{7} \)
Cross-multiplying gives:
\[ 2 \cdot 7 \neq -3 \cdot m \implies 14 \neq -3m \implies m \neq -\frac{14}{3} \]

### Conclusion
The values of \( m \) for which the system of equations has a unique solution are:
\[ m \neq 10 \quad \text{and} \quad m \neq -\frac{14}{3} \]