Obtain the condition for the following system of linear equations to has a unique solution
`px + qy = r` and `lx + my = n`.

Options

A. pm = ql
B. pm - ql = 0
C. pm `ne` ql
D. none of these

Answer

To determine the condition for the system of linear equations to have a unique solution, we start with the given equations:

1. \( px + qy = r \)
2. \( lx + my = n \)

### Step 1: Rewrite the equations in standard form
We can rewrite the equations in the standard form \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \).

- The first equation can be rewritten as:
\[ px + qy - r = 0 \]
Here, \( a_1 = p \), \( b_1 = q \), and \( c_1 = -r \).

- The second equation can be rewritten as:
\[ lx + my - n = 0 \]
Here, \( a_2 = l \), \( b_2 = m \), and \( c_2 = -n \).

### Step 2: Apply the condition for unique solutions
For a system of two linear equations to have a unique solution, the following condition must hold:
\[ \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \]

### Step 3: Substitute the values
Substituting the values we found:
\[ \frac{p}{l} \neq \frac{q}{m} \]

### Step 4: Cross-multiply to eliminate the fractions
Cross-multiplying gives us:
\[ pm \neq ql \]

### Conclusion
Thus, the condition for the system of linear equations \( px + qy = r \) and \( lx + my = n \) to have a unique solution is:
\[ pm \neq ql \]

---