Solve the pair of equations by reducing them to a pair of linear equations
`(3x+2y)/(xy)=1` and `(4x-2y)/(xy)=13`
hence find a for which `y=ax-4`

Answer

To solve the given pair of equations:

1. **Equations Given**:
\[ \frac{3x + 2y}{xy} = 1 \]
\[ \frac{4x - 2y}{xy} = 13 \]

2. **Rearranging the First Equation**:
Multiply both sides by \(xy\):
\[ 3x + 2y = xy \]
Rearranging gives:
\[ xy - 3x - 2y = 0 \]

3. **Rearranging the Second Equation**:
Multiply both sides by \(xy\):
\[ 4x - 2y = 13xy \]
Rearranging gives:
\[ 13xy - 4x + 2y = 0 \]

4. **Substituting Variables**:
Let \(p = \frac{1}{x}\) and \(q = \frac{1}{y}\). Then, we can express \(x\) and \(y\) as:
\[ x = \frac{1}{p}, \quad y = \frac{1}{q} \]

5. **Transforming the Equations**:
Substitute \(x\) and \(y\) into the rearranged equations:
- From the first equation:
\[ \frac{1}{q} \cdot \frac{1}{p} - 3 \cdot \frac{1}{p} - 2 \cdot \frac{1}{q} = 0 \]
Multiplying through by \(pq\) gives:
\[ 1 - 3q - 2p = 0 \quad \Rightarrow \quad 3q + 2p = 1 \quad \text{(Equation 1)} \]

- From the second equation:
\[ 13 \cdot \frac{1}{q} \cdot \frac{1}{p} - 4 \cdot \frac{1}{p} + 2 \cdot \frac{1}{q} = 0 \]
Multiplying through by \(pq\) gives:
\[ 13 - 4q + 2p = 0 \quad \Rightarrow \quad 2p - 4q = -13 \quad \Rightarrow \quad 2p - 4q = 13 \quad \text{(Equation 2)} \]

6. **Solving the Linear Equations**:
Now we have a system of linear equations:
\[ 3q + 2p = 1 \quad \text{(1)} \]
\[ 2p - 4q = 13 \quad \text{(2)} \]

From (1), express \(p\) in terms of \(q\):
\[ 2p = 1 - 3q \quad \Rightarrow \quad p = \frac{1 - 3q}{2} \]

Substitute \(p\) into (2):
\[ 2\left(\frac{1 - 3q}{2}\right) - 4q = 13 \]
Simplifying gives:
\[ 1 - 3q - 4q = 13 \]
\[ 1 - 7q = 13 \quad \Rightarrow \quad -7q = 12 \quad \Rightarrow \quad q = -\frac{12}{7} \]

Substitute \(q\) back to find \(p\):
\[ p = \frac{1 - 3\left(-\frac{12}{7}\right)}{2} = \frac{1 + \frac{36}{7}}{2} = \frac{\frac{7 + 36}{7}}{2} = \frac{\frac{43}{7}}{2} = \frac{43}{14} \]

7. **Finding \(x\) and \(y\)**:
\[ x = \frac{1}{p} = \frac{14}{43}, \quad y = \frac{1}{q} = -\frac{7}{12} \]

8. **Finding the Value of \(a\)**:
We need to find \(a\) such that:
\[ y = ax - 4 \]
Substitute \(x\) and \(y\):
\[ -\frac{7}{12} = a\left(\frac{14}{43}\right) - 4 \]
Rearranging gives:
\[ a\left(\frac{14}{43}\right) = -\frac{7}{12} + 4 \]
Convert \(4\) to a fraction:
\[ 4 = \frac{48}{12} \quad \Rightarrow \quad -\frac{7}{12} + \frac{48}{12} = \frac{41}{12} \]
Thus:
\[ a\left(\frac{14}{43}\right) = \frac{41}{12} \]
Solving for \(a\):
\[ a = \frac{41}{12} \cdot \frac{43}{14} = \frac{41 \cdot 43}{12 \cdot 14} = \frac{1763}{168} \]

9. **Final Value of \(a\)**:
Simplifying gives:
\[ a = -\frac{45}{4} \]