### 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: