Sistema

{(x + 2^(-1))^2 + (y + 2^(-1))^2 = x^2 + y^2

{x + 8^(- 1/3)·y = 4^(- 3/2)

-------------------------------------------

{(x + 1/2)^2 + (y + 1/2)^2 = x^2 + y^2

{x + 1/2·y = 1/8

-----------------------------------------------

{(x^2 + x + 1/4) + (y^2 + y + 1/4) = x^2 + y^2

{8·x + 4·y = 1

---------------------------------

{x + y + 1/2 = 0

{8·x + 4·y = 1

-----------------------------

risolvo per sostituzione: y = -x - 1/2

8·x + 4·(-x - 1/2) = 1

4·x - 2 = 1----> x = 3/4

y = - 3/4 - 1/2----> y = - 5/4

soluzione sistema: 

[x = 3/4 ∧ y = - 5/4]

============================================================

$\small \begin{Bmatrix}
\left(x+2^{-1}\right)^2+\left(y+2^{-1}\right)^2&=&x^2+y^2\\
x+8^{-\frac{1}{3}}y&=&4^{-\frac{3}{2}}\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
\left(x+\dfrac{1}{2}\right)^2+\left(y+\dfrac{1}{2}\right)^2&=&x^2+y^2\\
x+\left(\dfrac{1}{8}\right)^{\frac{1}{3}}y&=&\left(\dfrac{1}{4}\right)^{\frac{3}{2}}\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
\cancel{x^2}+x+\dfrac{1}{4}\cancel{+y^2}+y+\dfrac{1}{4}\cancel{-x^2}\cancel{-y^2}&=&0\\
x+\sqrt[3]{\dfrac{1}{8}}·y&=&\sqrt{\left(\dfrac{1}{4}\right)^3}\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
x+y&=&-\dfrac{1}{2}\\
x+\dfrac{1}{2}y&=&\sqrt{\dfrac{1}{64}}\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
x+y&=&-\dfrac{1}{2}\\
x+\dfrac{1}{2}y&=&\dfrac{1}{8}\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
x&=&-\dfrac{1}{2}-y\\
-\dfrac{1}{2}-y+\dfrac{1}{2}y&=&\dfrac{1}{8}\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
x&=&-\dfrac{1}{2}-y\\
-y+\dfrac{1}{2}y&=&\dfrac{1}{8}+\dfrac{1}{2}\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
x&=&-\dfrac{1}{2}-y\\
-8y+4y&=&1+4\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
x&=&-\dfrac{1}{2}-y\\
-4y&=&5\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
x&=&-\dfrac{1}{2}-y\\
4y&=&-5\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
x&=&-\dfrac{1}{2}-y\\
y&=&-\dfrac{5}{4}\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
x&=&-\dfrac{1}{2}-\left(-\dfrac{5}{4}\right)\\
y&=&-\dfrac{5}{4}\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
x&=&-\dfrac{1}{2}+\dfrac{5}{4}\\
y&=&-\dfrac{5}{4}\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
x&=&\dfrac{-2+5}{4}\\
y&=&-\dfrac{5}{4}\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
x&=&\dfrac{3}{4}\\
y&=&-\dfrac{5}{4}\\
\end{Bmatrix}$