I am aware that most of the times, it is difficult to explicitly find all equilibria of a nonlinear system by solving the system. I have the following parameterised nonlinear system, with seven nonnegative variables $x_1,x_2,...,x_7$.
$a,b,c,...,t$ are positive parameters
$\begin{cases}\dfrac{dx_1}{dt}=a-\dfrac{bcx_1x_2}{x_1+x_2}-cx_1\\ \dfrac{dx_2}{dt}=\dfrac{bcx_1x_2}{x_1+x_2}-cx_2\\ \dfrac{dx_3}{dt}=gd-\dfrac{efx_3x_4}{d}-\dfrac{hx_3x_7}{i+x_7}-gx_3\\ \dfrac{dx_4}{dt}=\dfrac{efx_3x_4}{d}+\dfrac{hx_3x_7}{i+x_7}-fx_4\\ \dfrac{dx_5}{dt}=j-\dfrac{kx_5x_4}{d}-\dfrac{lx_5x_2}{x_1+x_2}-\dfrac{m(o+n)rx_5x_6}{o(x_5+x_6(1+\frac{r}{o}+\frac{q}{n})}-\dfrac{px_5x_7}{i+x_7}-nx_5\\ \dfrac{dx_6}{dt}=\dfrac{kx_5x_4}{d}+\dfrac{lx_5x_2}{x_1+x_2}+\dfrac{m(o+n)rx_5x_6}{o(x_5+x_6(1+\frac{r}{o}+\frac{q}{n})}+\dfrac{px_5x_7}{i+x_7}-\dfrac{(o+nr)x_6}{0}\\ \dfrac{dx_7}{dt}=sx_2-tx_7\end{cases}$
When trying to find all equilibrium points by equating all equations to zero and solving, From the second equation, I have that $x_2=0$ or $x_1=\dfrac{x_2}{b-1}$. The case where $x_2=0$ is not the problem. Substituting $x_1=\dfrac{x_2}{b-1}$ into the first equation we have $x_2=\dfrac{a(b-1)}{cb}$.
Substituting $x_2$ into the seventh equation, we get $x_7=\dfrac{sa(b-1)}{cbt}$. Substituting $x_7$ into the third and the fourth equations, and the solving for $x_3$ in both, and equating the $x_3$s, I get a quadratic in $x_4$. This quadratic is so ugly, taking each root and substituting it into the remaining two equations is sooo messy. Is there an easier way to find the equilibrium points or at least be able to tell the number of equilibrium points and conditions under which they exist?