<s>[INST] Solve the system of equations $\sqrt{x+y}+\sqrt{2x+y+2}=7, 3x+2y=23$. Solve the system of equations $\sqrt{x+y}+\sqrt{2x+y+2}=7, 3x+2y=23$.
$x=\frac{23}{3}-\frac{2}{3} y$ so when I plug it in I get 
$\sqrt{\frac{23}{3}-\frac{2}{3}y+y} +\sqrt{2\times\frac{23}{3}-\frac{2}{3}y+y+2}=7$
$\sqrt{\frac{23}{3}+\frac{y}{3}} + \sqrt{\frac{52}{3}-\frac{y}{3}}=7$
$2\sqrt{\frac{1196}{9}+\frac{29}{9}y-\frac{y^{2}}{9}} = 24$
$\frac{1196}{9}+\frac{29}{9}y-\frac{y^{2}}{9}=144$
$y^{2}-29y+100=0$ 
$y_1=25, x_1=-9$ or $y_2=4, x_2=5$.
The solution is correct, but is everything written properly? Should I put $\iff$ at the beginning of each row? 
 [/INST] Put $a=x+y,\; b=2x+y+2$.
Equations:
$$\begin{cases}\begin{align}
&\sqrt{a}+\sqrt{b}=7&\Rightarrow a+b+2\sqrt{ab}&=49\\
&a+b-2=23&\Rightarrow a+b&=25\end{align}\bigg\rbrace\Rightarrow 25+2\sqrt{ab}=49\Rightarrow ab=144\\
\end{cases}\\$$
$$\begin{align}
a+\frac {144}a&=25\\
a^2-25a+144&=0\\
(a-16)(a-9)&=0\\
a&=16, 9\\
b&=9, 16\\
x=b-a-2&=-9,5\\
y=a-x&=25,4\\
\Rightarrow (x,y)&=(-9,25), (5,4)\;\blacksquare\end{align}$$
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