Given the following system:
$$ a_1+(a_3+a_4)V-a_5WX = 0 \\ (a_7+a_4)Y-a_6WZ = 0 \\ a_2-W(a_4+a_5X+a_6Z)+(a_3+a_8)V+a_7Y = 0 \\ a_5WX-(a_3+a_8+a_4)V = 0 \\ a_6WZ-(a_7+a_4)Y = 0 $$
where $a_1$ to $a_8$ are positive constants and $V,W,X,Y,Z$ are positive-valued variables, is it possible to solve by hand for $V,W,X,Y,Z$ in terms of the constants?
The only nonlinear terms in the system are $WX$ and $WZ$, and $X$ always occurs with $W.$
I have tried many things to solve this (i.e., Matlab and Maple), but am not getting explicit solutions for all, even though I have $5$ equations in $5$ variables. I always get one free variable.
I would appreciate help.
Your second and fifth equation are equivalent. This might be immediately clear from the first display below.
From $$\begin{align} a_1 + a_3 V + a_4 V - a_5 WX &= 0 & (1) \\ a_4 Y + a_7 Y - a_6 WZ &= 0 & (2) \\ a_2 + a_3 V + a_8 V - a_4 W - a_5 W + a_7 Y - a_6 WZ &= 0 & (3) \\ -a_3 V - a_4 V - a_8 V + a_5 WX &= 0 &(4) \\ -a_4 Y - a_7 Y + a_6 WZ &= 0 & (5) \end{align}$$ eliminate $WX$ and simplify some obvious things, yielding $$\begin{align} a_1 &= a_8 V & \text{ from (1) and (4)}\\ a_2 &= -a_3 V - a_8 V + a_4 W + a_5 W + a_4 Y & \text{ (3) + (5) }\\ a_6 WZ &= (a_4 + a_7) Y & \text{ from (2) or (5) } \end{align}$$
We're down to three equations in four unknowns, so there will be a free variable. Also, this system is extremely simple to finish by hand.