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Solving the first equation for $x_2$ yields $$ x_{{2}}={\frac {{x_{{1}}}^{3}}{2\;a_{{1}}}}+3\,{\frac {{x_{{1}}}^{2}}{a_{{1} }}}-{\frac { \left( a_{{1}}-4 \right) x_{{1}}}{a_{{1}}}}-4 $$ This cubic is $0$ at $x_1 = -4$ and $-1 \pm \sqrt{1+2 a_1}$. It is convex and increasing for $x_1 > -1 + \sqrt{1+2 a_1}$. Similarly, for the second equation, with indices $1$ and $2$ interchanged. From this it is easy to see that the two curves intersect exactly once in the first quadrant.

Solving the first equation for $x_2$ yields $$ x_2 = \frac{3 x_1^2}{2 a_1} + \frac{6 x_1}{a_1} + \frac{4}{a_1} - 1$$ This is $0$ at $x_1 = \sqrt{1+2 a_1} - 1$, increasing to the right of that, and convex, with $x_2/x_1 \to +\infty$ as $x_1 \to +\infty$. Similarly with indices reversed for the second equation. It is then easy to see that there is a solution with $x_1, x_2 > 0$, and that this is unique.

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