Theory:

We will learn how to find the solutions to the pair of equations that are not linear but are reducible to a pair of linear equations.
 
Let us discuss the solution of such a system of equation using an example.
Example:
Consider the following pair of equations:
 
\(2x + 3y = 5xy\) ……\((1)\)
 
\(x + 3y = 7xy\)   ……\((2)\)
 
Let simplify the equations \((1)\) and \((2)\).
 
Divide both the equations throughout by \(xy\).
 
\(\frac{2}{y} + \frac{3}{x} = 5\) ……\((3)\)
 
\(\frac{1}{y} + \frac{3}{x} = 7\) ……\((4)\)
 
Rearrange and rewrite the equations by separating the variables.
 
\(3\left(\frac{1}{x}\right) + 2\left(\frac{1}{y}\right) = 5\) ……\((5)\)
 
\(3\left(\frac{1}{x}\right) + \left(\frac{1}{y}\right) = 7\)   ……\((6)\)
 
Now, let us transform the equation to the linear form \(ax + by + c = 0\) by making a relevant substitution.
 
Substitute \(\frac{1}{x}\) \(=\) \(a\) and \(\frac{1}{y}\) \(=\) \(b\) in the equations \((5)\) and \((6)\).
 
\(3a + 2b = 5\) ……\((7)\)
 
\(3a + b = 7\)   ……\((8)\)
 
Solve the above two equations by one of the three solving methods.
 
We will solve the equations \((7)\) and \((8)\) by the elimination method.
 
Subtract equation \((8)\) from \((7)\).
 
We have \(b\) \(=\) \(-2\).
 
Substitute \(b\) \(=\) \(-2\) in equation \((8)\).
 
\(\Rightarrow\) \(3a + (-2) = 7\)
 
\(\Rightarrow\) \(3a - 2 = 7\)
 
\(\Rightarrow\) \(3a = 7 + 2\)
 
\(\Rightarrow\) \(3a = 9\)
 
\(\Rightarrow\) \(a = \frac{9}{3}\)
 
\(\Rightarrow\) \(a = 3\)
 
Thus \(a = 3\) and \(b = -2\).
 
Here, \(\frac{1}{x}\) \(=\) \(a\).
 
\(\frac{1}{x}\) \(=\) \(3\)
 
This implies \(x\) \(=\) \(\frac{1}{3}\)
 
Similarly, \(\frac{1}{y}\) \(=\) \(b\).
 
Here, \(\frac{1}{y}\) \(=\) \(b\).
 
\(\frac{1}{y}\) \(=\) \(-2\)
 
This implies \(y\) \(=\) \(-\frac{1}{2}\)
 
Therefore the solution to the pair of equations \(2x + 3y = 5xy\) and \(x + 3y = 7xy\) is \(\left(\frac{1}{3}, -\frac{1}{2}\right)\).
 
Verification:
 
\(2\left(\frac{1}{3}\right) + 3\left(-\frac{1}{2}\right) = 5\left(\frac{1}{3}\right)\left(-\frac{1}{2}\right)\)
 
\(\frac{2}{3} - \frac{3}{2} = 5\left(-\frac{1}{6}\right)\)
 
\(-\frac{5}{6} = -\frac{5}{6}\)
 
Hence, verified.
Important!
The three methods of solving the pair of simultaneous linear equations are: