Solve the following system of equations: $\;$ $\sqrt{x} + \sqrt{y} = 10$, $\;$ $\sqrt[4]{x} + \sqrt[4]{y} = 4$
Given system of equations:
$\sqrt{x} + \sqrt{y} = 10$ $\;\;\; \cdots \; (1)$, $\;\;$ $\sqrt[4]{x} + \sqrt[4]{y} = 4$ $\;\;\; \cdots \; (2)$
Let $\;\;$ $\sqrt[4]{x} = p$ $\;\;\; \cdots \; (3a)$, $\;$ $\sqrt[4]{y} = q$ $\;\;\; \cdots \; (3b)$
In view of equations $(3a)$ and $(3b)$, equations $(1)$ and $(2)$ respectively become
$p^2 + q^2 = 10$ $\;\;\; \cdots \; (1a)$ $\;\;$ and $\;\;$ $p + q = 4$ $\;\;\; \cdots \; (2a)$
We have from equation $(2a)$, $\;\;$ $p = 4 - q$ $\;\;\; \cdots \; (4)$
In view of equation $(4)$, equation $(1a)$ becomes
$\left(4 - q\right)^2 + q^2 = 10$
i.e. $\;$ $16 - 8q + q^2 + q^2 = 10$
i.e. $\;$ $2q^2 - 8q + 6 = 0$
i.e. $\;$ $q^2 - 4q + 3 = 0$
i.e. $\;$ $\left(q - 3\right) \left(q - 1\right) = 0$
i.e. $\;$ $q = 3$ $\;\;\;$ or $\;\;\;$ $q = 1$
When $\;$ $q = 3$, $\;$ we have from equation $(4)$, $\;$ $p = 4 - 3 = 1$
When $\;$ $q = 1$, $\;$ we have from equation $(5)$, $\;$ $p = 4 - 1 = 3$
For $\;$ $p = 1, \; q = 3$, $\;$ we have from equations $(3a)$ and $(3b)$
$\sqrt[4]{x} = 1$ $\implies$ $x = 1$ $\;\;$ and $\;\;$ $\sqrt[4]{y} = 3$ $\implies$ $y = 81$
For $\;$ $p = 3, \; q = 1$, $\;$ we have from equations $(3a)$ and $(3b)$
$\sqrt[4]{x} = 3$ $\implies$ $x = 81$ $\;\;$ and $\;\;$ $\sqrt[4]{y} = 1$ $\implies$ $y = 1$
$\therefore \;$ The solution to the given system of equations is $\left(x, y\right) = \left\{\left(1, 81\right), \left(81, 1\right) \right\}$