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Is it possible to determine real solutions $x,y$ for the following system of nonlinear equations in terms of the real constants $a,b,k$?

$$ a=kx(1-y^2)\qquad\mbox{and}\qquad b=ky(1-x^2). $$ If so, what is (are) the solution(s)?

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You can try out a graphical solution here: link

enter image description here

The system has been transformed to $$ A = x(1-y^2) \quad \quad B = y (1-x^2) $$ where $A = a/k$ and $B= b/k$.

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Square the first equation $a^2 = k^2x^2(1-y^2)^2$. Now $k^2(1-y^2)^2$ minus this equation gives \begin{eqnarray*} k^2(1-y^2)^2-a^2 = k^2(1-y^2)^2(1-x^2) \end{eqnarray*} Multiply this by $y$ and substitute for $ky(1-x^2)=b$ and we have \begin{eqnarray*} \color{red}{y(k^2(1-y^2)^2-a^2) = kb(1-y^2)^2}. \end{eqnarray*} This is a quintic (order $5$) in $y$ and will not be easy to solve.

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