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$$\frac { - \left( 2 e ^ { h + l } \left( 2 ( - c + l - 1 ) e ^ { h + l + l } + ( n - 2 ) ( - c + l - 1 ) e ^ { h + l } + n ( - c + l - 1 ) e ^ { l + q + l } - 2 e ^ { h + l } \right) \right) } { \left( 2 e ^ { h + l + l } + 2 e ^ { h + l } + ( n - 2 ) e ^ { h + l } + n e ^ { d + q + l } \right) ^ { 2 } } = 0$$

and

$$\frac { \left( 2 e ^ { h + l + q } \left( 2 ( c - h + 1 ) e ^ { h + l + h } + n ( c - h + 1 ) e ^ { h + h } - ( n - 2 ) ( - c + h - 1 ) e ^ { l + q + h } + 2 e ^ { + l + q } \right) \right) } { \left( 2 e ^ { h + l + q } + 2 e ^ { h + l + h } + n e ^ { h + l } + ( n - 2 ) e ^ { l + q + h } \right) ^ { 2 } } = 0$$

The parameters are $q , n ,$ and the variables are $h$ and $l$. All parameters and variables are strictly positive. I am looking for a set of conditions on the parameters, such that:

  1. $h , l > c$
  2. $n > 2$
  3. $q - h > - l$
  4. Optional $: h > l$ (I have a feeling that this condition should hold)

Question

All I want is a hint as to how to proceed -- I'm not asking for an answer, just a hint.

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    $\begingroup$ The first thing I would do is multiply both sides of each equation by the denominator on the left. $\endgroup$
    – user247327
    Feb 8, 2019 at 12:13
  • $\begingroup$ Note that $\frac ab=0 \implies a=0\cdot b \implies a=0$ $\endgroup$
    – lioness99a
    Feb 8, 2019 at 12:22

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