Answer
$(-\sqrt 2, \sqrt 2)$
$(-\sqrt 2, -\sqrt 2)$
$(\sqrt 2, \sqrt 2)$
$(\sqrt 2, -\sqrt 2)$
Work Step by Step
Given the system of equations:
1. $\frac{4}{x^2} + \frac{6}{y^4} = 3.5$
2. $\frac{1}{x^2} - \frac{2}{y^4} = 0$
It asks to find all solutions.
Triple equation two, add the result to equation 1
$\frac{4}{x^2} + \frac{6}{y^4} = 3.5$
+($\frac{3}{x^2} - \frac{6}{y^4} = 0$)
$\frac{7}{x^2} = 3.5$
$x^2 = 2$
$x = +/- \sqrt 2$
Substitute the values into the equation
($\frac{3}{\sqrt 2^2} - \frac{6}{y^4} = 0$)
$\frac{3}{2} = \frac{6}{y^4}$
$y^4 = 4$
$y = +/- \sqrt 2$
Same solution for $x = -\sqrt 2$
All solutions:
$(-\sqrt 2, \sqrt 2)$
$(-\sqrt 2, -\sqrt 2)$
$(\sqrt 2, \sqrt 2)$
$(\sqrt 2, -\sqrt 2)$
