Task B-2.3. Solve the system of equations:

$$
\begin{aligned}
& x+y+\sqrt{x+y}=6 \\
& x-y+\sqrt{x-y}=2
\end{aligned}
$$
## Solution.

Let $x+y=a, x-y=b$. The given system then transforms into

$$
\begin{aligned}
& a+\sqrt{a}=6 \\
& b+\sqrt{b}=2
\end{aligned}
$$

Solve the first equation of the system.

$$
\begin{aligned}
& \sqrt{a}=6-a /^{2}, \text { with the condition } 6-a \geq 0 \\
& a^{2}-13 a+36=0 \\
& a_{1,2}=\frac{13 \pm \sqrt{13^{2}-4 \cdot 36}}{2} \\
& a_{1,2}=\frac{13 \pm 5}{2} \\
& a_{1}=9, a_{2}=4
\end{aligned}
$$

The only solution $a=4$ satisfies the given condition.

(1 point)

Solve the second equation of the system:

$$
\begin{aligned}
& \sqrt{b}=2-b /^{2}, \text { with the condition } 2-b \geq 0 \\
& b^{2}-5 b+4=0 \\
& b_{1}=4, b_{2}=1
\end{aligned}
$$

After checking the conditions, the only solution that remains is $b=1$.

(1 point)

It remains to solve the system

$$
\left\{\begin{array}{l}
x+y=4 \\
x-y=1
\end{array}\right.
$$

The solution to the system is $(2.5,1.5)$.