Solve the following system of equations: $\;$ $x + y = 7, \;\; y = \dfrac{6}{x}$
Given system of equations:
$x + y = 7$ $\;\;\; \cdots \; (1)$
$y = \dfrac{6}{x}$ $\;\;\; \cdots \; (2)$
In view of equation $(2)$, equation $(1)$ becomes
$x + \dfrac{6}{x} = 7$
i.e. $\;$ $x^2 - 7x + 6 = 0$
i.e. $\;$ $\left(x - 6\right) \left(x - 1\right) = 0$
i.e. $\;$ $x = 6$ $\;$ or $\;$ $x = 1$
Substituting the value of $x$ in equation $(2)$ gives
when $\;$ $x = 6$, $\;$ $y = \dfrac{6}{6} = 1$
and when $\;$ $x = 1$, $\;$ $y = \dfrac{6}{1} = 6$
$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(6, 1\right), \left(1, 6\right) \right\}$