Of course what you asked for is information about computer systems for solving this kind of thing, but the particular system of equations you have here isn't so hard to make sense of (but, caution, I make a lot of mistakes, so if you want to make any use of what follows I strongly advise you to check it). So, let's begin with the quantities z,m,n,b,c. z is known; m,n are constrained by having to add up to z, and z,m,n interact with the other variables only via b,c. Oh, and we also know y=b+c, so we've got two constraints on b,c which means we should expect there to be only finitely many possibilities for (b,c); let's see what they are. This sort of thing is a bit painful by hand, but e.g. Mathematica can do it instantly. (Though it doesn't give what seems to me the simplest form, so I've made some hand-simplifications which of course might have introduced errors.) It turns out that b,c are  y2±z2√y2−5z2y2−z2 which isn't too painful. OK, now we split up x as q+r and l as s+t, and then from b,q we calculate o and from c,s we calculate v, and then we have two expressions for w which need to be equal. So that gives us b2−q2+r2=c2−s2+t2 which looks a bit unpleasant, but notice that we can write r2−q2=(r−q)x where r−q can take any value in [−x,+x] and similarly for s,t,l. We have b2−c2=±yz√y2−5z2y2−z2 and this has to equal q2−r2+t2−s2 which by the remarks in the previous paragraph is δx+ϵl where |δ|≤x and|ϵ|≤l and can therefore take any value between 0 and x2+l2. So, first of all, the whole thing has solutions iff yz√y2−5z2y2−z2≤x2+l2. And now we get w2 by offsetting b2 by something between ±x2, and also by offsetting c2 by something between ±l2, and so the smallest we can make w2 is max(0,b2−x2,c2−l2). (But remember that we can take b,c whichever way around we prefer; so we should put the + sign in b if x>l and in c if x<l.)