ĐK: $\left\{\begin{array}{l}4x+5y\ge0\\3x+y\ge0\end{array}\right.$
Đặt: $u=\sqrt{4x+5y};v=\sqrt{3x+y};u,v\ge0$, hệ trở thành:
$\left\{\begin{array}{l}u+v=5\\2v+\dfrac{7}{11}u^2+\dfrac{53}{11}v^2=29\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}u=5-v\\2v+\dfrac{7}{11}(5-v)^2+\dfrac{53}{11}v^2=29\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}u=5-v\\\dfrac{60}{11}v^2-\dfrac{48}{11}v-\dfrac{144}{11}=0\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}v=2\\u=3\end{array}\right.$ (vì $u,v\ge0$)
$\Leftrightarrow \left\{\begin{array}{l}4x+5y=9\\3x+y=4\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}x=1\\y=1\end{array}\right.$