Solve the following system of equations: $\;$ $\left(x^2 + y^2\right)xy = 78, \;\; x^4 + y^4 = 97$
Given system of equations:
$\left(x^2 + y^2\right) xy = 78$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $x^4 + y^4 = 97$ $\;\;\; \cdots \; (2)$
Equation $(2)$ can be written as
$\left(x^2 + y^2\right)^2 - 2 x^2 y^2 = 97$ $\;\;\; \cdots \; (2a)$
Let $\;\;$ $x^2 + y^2 = u$ $\;\;\; \cdots \; (3a)$
and $\;\;$ $xy = v$ $\;\;\; \cdots \; (3b)$ $\;\;\;$ i.e. $\;$ $x = \dfrac{v}{y}$ $\;\;\; \cdots \; (3c)$
In view of equations $(3a)$ and $(3b)$, equations $(1)$ and $(2a)$ become
$uv = 78$ $\;\;\;$ i.e. $\;$ $u = \dfrac{78}{v}$ $\;\;\; \cdots \; (4)$
and $\;\;$ $u^2 - 2v^2 = 97$ $\;\;\; \cdots \; (5)$
Substituting for $u$ from equation $(4)$ in equation $(5)$ gives
$\left(\dfrac{78}{v}\right)^2 - 2v^2 = 97$
i.e. $\;$ $6084 - 2v^4 = 97v^2$
i.e. $\;$ $2v^4 + 97v^2 - 6084 = 0$
i.e. $\;$ $\left(2v^2 + 169\right)$ $\left(v^2 - 36\right) = 0$
i.e. $\;$ $v^2 = \dfrac{-169}{2} = - 84.5$ $\;\;$ or $\;\;$ $v^2 = 36$
Since $v^2$ cannot be negative, the only possible solution is
$v^2 = 36$ $\;\;\;$ i.e. $\;$ $v = \pm 6$
Substituting the value of $v$ in equation $(4)$ gives
when $\;$ $v = +6$, $\;\;$ $u = \dfrac{78}{6} = 13$ $\;\;$ and when $\;$ $v = -6$, $\;\;$ $u = \dfrac{78}{-6} = - 13$
Substituting the values of $u$ and $v$ in equations $(3a)$ and $(3c)$ gives
- when $\;\;$ $u = 13, \; v = 6$, $\;\;$ we have
$x^2 + y^2 = 13$ $\;\;\; \cdots \; (6)$ $\;\;$ and $\;\;$ $x = \dfrac{6}{y}$ $\;\;\; \cdots \; (7)$
In view of equation $(7)$, equation $(6)$ becomes
$\left(\dfrac{6}{y}\right)^2 + y^2 = 13$
i.e. $\;$ $y^4 - 13y^2 + 36 = 0$
i.e. $\;$ $\left(y^2 - 9\right) \left(y^2 - 4\right) = 0$
i.e. $\;$ $y^2 = 9$ $\;\;$ or $\;\;$ $y^2 = 4$
i.e. $\;$ $y = \pm 3$ $\;\;$ or $\;\;$ $y = \pm 2$
Substituting the value of $y$ in equation $(7)$ gives
when $\;\;$ $y = \pm 3$, $\;\;$ $x = \dfrac{6}{\pm 3} = \pm 2$
and when $\;\;$ $y = \pm 2$, $\;\;$ $x = \dfrac{6}{\pm 2} = \pm 3$ - when $\;\;$ $u = -13, \; v = -6$, $\;\;$ we have
$x^2 + y^2 = -13$ $\;\;\; \cdots \; (8)$ $\;\;$ and $\;\;$ $x = \dfrac{-6}{y}$ $\;\;\; \cdots \; (9)$
Since the sum of squares of any two numbers cannot be negative, $\;$ $u = -13, \; v = -6$ $\;$ are not valid solutions.