Denote $x=\frac{t+u}{2},\ y=\frac{t-u}{2}$, where $t>u$. Then $x+y=t$ (obviously $t>0$) and we can solve the system in terms of $t,u$. Notice that if $(t,u)$ is a solution, then we must have $$\begin{cases}\left(2+\frac{5}{t}\right)^2=\frac{18}{t+u}\\ \left(2-\frac{5}{t}\right)^2=\frac{8}{t-u}\end{cases}$$meaning that also $$\left(2+\frac{5}{t}\right)^2-\left(2-\frac{5}{t}\right)^2=\frac{18}{t+u}-\frac{8}{t-u}$$or $$\frac{20}{t}=\frac{9}{t+u}-\frac{4}{t-u}$$Hence the only possibility to check: $u=-\frac{3}{5}t$, because $u=\frac{5}{4}t$ does not verify $t>u$ when $t>0$. That yields $t=\frac{5}{4}$ or $t=5$ of which it is easy to choose the greater and check that together with $u=-3$ it is indeed a solution.

jgamer2768 wrote: Let $x$ and $y$ be positive real numbers satisfying the following system of equations: $\begin{cases}\sqrt{x}\left(2 + \dfrac{5}{x+y}\right) = 3 \\\\ \sqrt{y}\left(2 - \dfrac{5}{x+y}\right) = 2 \\\end{cases}$ Find the maximum value of $x$ $+$ $y$. Already posted two days ago : https://artofproblemsolving.com/community/c6h3421990p32957074

Solved with Jorge Liu. The answer is $x+y=$ 5. First, $x+y=5$ is possible if $x=1$ and $y=4$. Now, let’s prove that it is the maximum. Assume for contradiction that there exist $x, y$ that satisfy the system of equations such that $x+y>5$. If this is the case, by the second equation, we have that \[ \sqrt{y}=\sqrt{y}\left(2-\frac{5}{5}\right)<\sqrt{y}\left(2-\frac{5}{x+y}\right)=2, \]so $y<4$. Now, we will prove that $x\leqslant 1$. Assume for contradiction that $x>1$. In this case, by the first equation, we have that \begin{align*} \sqrt{x}\left(2+\frac{5}{x+y}\right)&=3\\ \sqrt{x}\left(2+\frac{5}{x+4}\right)&<3\\ 2x\sqrt{x}+13\sqrt{x}&<3x+12\\ (2x+13)(\sqrt{x}-1)&<x-1\\ 2x+13&<\sqrt{x}+1\\ 2x+12&<\sqrt{x}, \end{align*}so that \[ \sqrt{x}<x<2x+12<\sqrt{x}, \]but this is a contradiction. Therefore, we must have that $x\leqslant 1$. However, this implies that \[ x+y<1+4=5, \]but we had assumed that $x+y>5$. This is a contradiction. Therefore, $x+y\leqslant5$, as we wanted. $\square$