Let two variables $x,y\in \mathbb{R}$ be defined by the curves \begin{align} x=\frac{-(ab+c)y^2+(a-m)y}{cy-m}\\ y=\frac{-(ef+d)x^2+(e-m)x}{dx-m} \end{align} Where $a,b,c,d,e,f,m\in\mathbb{R}$ and are all positive nonzero constants. Assume that $cy-m\not=0$, $dx-m\not=0$ and $x,y>0$. How many solutions are possible and what conditions on the coefficients are necessary to ensure that number of solutions?
I am confused what methods I can/should use to answer this question.
Notes
To clarify what the question we are asking if we instead replace the system with $ax^2+bx+c=0$, the answer would be
- If $a c<0$ there is one positive real root and one negative real root.
- If $ ac>0$,$b^2 -4 ac> 0$ and $\dfrac{-b}{2 a}>0$ then there are two real positive roots.
- If $ac>0$, $b^2 -4 ac=0$ and $\dfrac{-b}{2 a}>0$ there is one real positive root.
Note sure what the appropriate tags are for this question so please feel free to edit the tags.
- References to papers or texts would also be appreciated.
- I am not looking for the explicit form of the solutions but the number of solutions.
- If you need any clarification feel free to ask.
The particular form of these equations allows us to make some simplifications. Let $z=x/y$. The two equations are equivalent to $$\begin{eqnarray*} z&=&\frac{-(ab+c)y+(a-m)}{cy-m},\\ z&=&\frac{dx-m}{-(ef+d)x+(e-m)}. \end{eqnarray*}$$ We can rearrange to express $x,y$ in terms of $z$: $$\begin{eqnarray*} y&=&\frac{mz+(a-m)}{cz+(ab+c)},\\ x&=&\frac{(e-m)z+m}{(ef+d)z+d}. \end{eqnarray*}$$ (Note that each denominator is positive by assumption.) Now $z$ must satisfy the cubic equation $$ z(mz+(a-m))((ef+d)z+d)-(cz+(ab+c))((e-m)z+m)=0. $$ So our task is equivalent to counting how many roots of this cubic are greater than $\max(0,1-a/m)$. In particular there are at most $3$. Probably it is possible to find exact criteria in terms of the constants by using the discriminant of the cubic.