Chapter 5, Systems of Equations and Inequalities - Section 5.4 - Systems of Nonlinear Equations - 5.4 Exercises - Page 466: 32

$(\pm\sqrt2,\pm\sqrt2)$

Let $1/x^2=a,1/y^4=b$ Then our system of equations is: $\{4a+6b=3.5, a-2b=0\}$. Adding three times the second equation to the first one, we get: $7a=3.5\\a=0.5$ Plugging this back into the first equation, we get: $4(0.5)+6b=3.5\\6b=1.5\\b=0.25$ $x^2=1/a=2\\x=\pm\sqrt2$ $y^4=1/b=4\\y=\pm\sqrt2$ Hence, the solutions are: $(\pm\sqrt2,\pm\sqrt2)$