Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $x^3 + y^3 = 35, \;\; x + y = 5$


Given system of equations:

$x^3 + y^3 = 35$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $x + y = 5$ $\;\;\; \cdots \; (2)$

We have from equation $(2)$, $\;\;\;$ $x = 5 - y$ $\;\;\; \cdots \; (3)$

In view of equation $(3)$, equation $(1)$ becomes

$\left(5 - y\right)^3 + y^3 = 35$

i.e. $\;$ $125 - 75y + 15 y^2 - y^3 + y^3 = 35$

i.e. $\;$ $15 y^2 - 75 y + 90 = 0$

i.e. $\;$ $y^2 - 5y + 6 = 0$

i.e. $\;$ $\left(y - 3\right) \left(y - 2\right) = 0$

i.e. $\;$ $y = 3$ $\;\;$ or $\;\;$ $y = 2$

$\therefore \;$ We have from equation $(3)$,

when $\;$ $y = 3$, $\;$ $x = 5 - 3 = 2$

and when $\;$ $y = 2$, $\;$ $x = 5 - 2 = 3$

$\therefore \;$ The solution to the given system of equations is $\;\;$ $\left(x, y\right) = \left\{\left(2, 3\right), \; \left(3, 2\right) \right\}$