JanHaj wrote: Find all real numbers $a$ and $b$ that satisfy the system of equations: $$\begin{cases} a &= \frac{2}{a+b} \\ \\ b &= \frac{2}{3a-b} \\ \end{cases}$$ Let $ab=t$ First equation is $a^2=2-t$ while second equation implies $b^2=3t-2$ So $t^2=a^2b^2=(2-t)(3t-2)$, which is $t^2-2t+1=0$ and so $t=1$ and $a^2=b^2=1$ And so $\boxed{(a,b)\in\{(-1,-1),(1,1)\}}$ which indeed both fit (the solutions $a=-b=\pm 1$ are forbiden by first equation, for example)

Simplify to get (a-b) ^2 = 0, I. E. a= b. From here the only sols are (1, 1) and (-1, -1).

From the system of equations we have that: 1) a^2+ab=2, and 2) 3ab-b^2=2, so basically a^2+ab=3ab-b^2=2. From a^2+ab=3ab-b^2 we have that (a-b)^2=0, so a=b. Then we just substitute b with a in the first equation, where we find that the solutions are (1,1) and (-1,-1).