Solve the following system of equations: $\;$ $x^2 + y^2 = 41, \;\; y - x = 1$
Given system of equations:
$x^2 + y^2 = 41$ $\;\;\; \cdots \; (1)$
$y - x = 1$ $\;$ i.e. $\;$ $y = x + 1$ $\;\;\; \cdots \; (2)$
In view of equation $(2)$, equation $(1)$ becomes
$x^2 + \left(x + 1\right)^2 = 41$
i.e. $\;$ $x^2 + x^2 + 2x + 1 = 41$
i.e. $\;$ $2x^2 + 2x - 40 = 0$
i.e. $\;$ $x^2 + x - 20 = 0$
i.e. $\;$ $\left(x + 5\right) \left(x - 4\right) = 0$
i.e. $\;$ $x = -5$ $\;$ or $\;$ $x = 4$
Substituting the value of $x$ in equation $(2)$ gives
When $\;$ $x = -5$, $\;$ $y = -5 + 1 = -4$
and when $\;$ $x = 4$, $\;$ $y = 4 + 1 = 5$
$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(-5, -4\right), \left(4, 5\right) \right\}$