5. Given real numbers $x, y$ satisfy
$$
\frac{4}{x^{4}}-\frac{2}{x^{2}}=3, y^{4}+y^{2}=3 \text {. }
$$

Then the value of $\frac{4}{x^{4}}+y^{4}$ is $\qquad$
(2008, "Mathematics Weekly Cup" National Junior High School Mathematics Competition)
By observation, we know that $-\frac{2}{x^{2}}$ and $y^{2}$ are two different solutions of the equation $m^{2}+m=3$.
By Vieta's formulas, we have
$$
\begin{array}{l}
\frac{4}{x^{4}}+y^{4}=m_{1}^{2}+m_{2}^{2}=\left(m_{1}+m_{2}\right)^{2}-2 m_{1} m_{2} \\
=(-1)^{2}-2(-3)=7 .
\end{array}
$$