Solve the following pair of linear equations :
`px + qy= p-q`
`qx- py= p+q`

Answer

To solve the pair of linear equations:

1. **Equations**:
\[ px + qy = p - q \quad \text{(1)} \]
\[ qx - py = p + q \quad \text{(2)} \]

2. **Multiply the first equation by \( q \)** and the second equation by \( p \):
\[ q(px + qy) = q(p - q) \implies pqx + q^2y = qp - q^2 \quad \text{(3)} \]
\[ p(qx - py) = p(p + q) \implies pqx - p^2y = p^2 + pq \quad \text{(4)} \]

3. **Now, we will subtract equation (4) from equation (3)**:
\[ (pqx + q^2y) - (pqx - p^2y) = (qp - q^2) - (p^2 + pq) \]
This simplifies to:
\[ q^2y + p^2y = qp - q^2 - p^2 - pq \]
\[ (q^2 + p^2)y = qp - q^2 - p^2 - pq \]

4. **Rearranging the right-hand side**:
\[ (q^2 + p^2)y = - (p^2 + q^2) \]
\[ y = \frac{- (p^2 + q^2)}{q^2 + p^2} = -1 \]

5. **Substituting \( y = -1 \) back into equation (1)** to find \( x \):
\[ px + q(-1) = p - q \]
\[ px - q = p - q \]
\[ px = p - q + q \]
\[ px = p \]
\[ x = \frac{p}{p} = 1 \]

6. **Final Solution**:
\[ x = 1, \quad y = -1 \]