Solve the following system of equations: $\;$ $x^4 + y^4 = 82, \;\; xy = 3$
Given system of equations:
$x^4 + y^4 = 82$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $xy = 3$ $\;\;\; \cdots \; (2)$
Equation $(1)$ can be written as
$\left(x^2\right)^2 + \left(y^2\right)^2 = 82$
i.e. $\;$ $\left(x^2 + y^2\right)^2 - 2 \left(xy\right)^2 = 82$
i.e. $\;$ $\left(x^2 + y^2\right)^2 - 18 = 82$ $\;\;\;$ [by equation $(2)$]
i.e. $\;$ $\left(x^2 + y^2\right)^2 = 100$
i.e. $\;$ $x^2 + y^2 = \pm 10$
- When $\;$ $x^2 + y^2 = 10$
i.e. $\;$ $\left(x + y\right)^2 - 2xy = 10$
i.e. $\;$ $\left(x + y\right)^2 - 6 = 10$ $\;\;\;$ [by equation $(2)$]
i.e. $\;$ $\left(x + y\right)^2 = 16$
i.e. $\;$ $x + y = \pm 4$ - When $\;$ $x + y = + 4$ $\;\;\;$ i.e. $\;$ $x = 4 - y$ $\;\;\; \cdots \; (3)$
we have from equation $(2)$,
$\left(4 - y\right) y = 3$
i.e. $\;$ $y^2 - 4y + 3 = 0$
i.e. $\;$ $\left(y - 3\right) \left(y - 1\right) = 0$
i.e. $\;$ $y = 3$ $\;\;$ or $\;\;$ $y = 1$
Substituting the values of of $y$ in equation $(3)$ gives - when $\;$ $y = 3$, $\;\;$ $x = 4 - 3 = 1$
- when $\;$ $y = 1$, $\;\;$ $x = 4 - 1 = 3$
- When $\;$ $x + y = - 4$ $\;\;\;$ i.e. $\;$ $x = -4 - y$ $\;\;\; \cdots \; (4)$
we have from equation $(2)$,
$- \left(4 + y\right) y = 3$
i.e. $\;$ $y^2 + 4y + 3 = 0$
i.e. $\;$ $\left(y + 3\right) \left(y + 1\right) = 0$
i.e. $\;$ $y = - 3$ $\;\;$ or $\;\;$ $y = -1$
Substituting the values of of $y$ in equation $(4)$ gives - when $\;$ $y = -3$, $\;\;$ $x = -4 - \left(-3\right) = -1$
- when $\;$ $y = -1$, $\;\;$ $x = -4 - \left(-1\right) = -3$
- When $\;$ $x^2 + y^2 = -10$
i.e. $\;$ $\left(x + y\right)^2 - 2xy = -10$
i.e. $\;$ $\left(x + y\right)^2 - 6 = -10$ $\;\;\;$ [by equation $(2)$]
i.e. $\;$ $\left(x + y\right)^2 = -16$
But square of any number cannot be negative.
$\therefore \;$ $x^2 + y^2 = -10$ $\;$ is not a valid solution.