@lt Your Reply is a completely new Question rather than a followup to the previous Answer. Please post it as a new Question.

@jockoflocko You wrote:

...I thought that every variable declared outside of a procedure could be used in the procedure without declaring it again....

That's mostly true. You can access the value of any variable declared in a scope more global than the procedure without needing to declare it again. But if the variable appears on the left side of an assignment statement, then it needs to be declared global.

@wolfman29 For a real range, the left endpoint can be -infinity and the right endpoint can be infinity. A complex range must be a rectangle (possibly infinite) specified by (lower left corner)..(upper right corner). For example:

Simply real: -infinity..infinity
A finite complex range: -1-I..1+I
The right half plane (real part nonnegative): 0-I*infinity..infinity+I*infinity

@Markiyan Hirnyk Yes. Considering that this will be done in floating point, a threshold volume to be considered nonplanar can be based on the setting of Digits. I only envision using this method for smooth curves. A nonplanar piecewise curve could contain a planar piece. If all the randomly selected points are on this planar piece, it defeats the algorithm.

@Axel Vogt 

I had mistakenly used 11+1/2 instead 11/2 to replace the 5.5 in the original integrand. After correcting that, I get the same answer as you and as Matlab. I have updated my Answer above.

@sami131 

Note that the expression being plotted has both c0 and m in denominators, so neither of those parameters can be 0. Other than that, putting the plots together is trivial. Indeed, even display is not needed, although it would also be trivial to use it.

M:= [1/2,1/3,3/4,2,3]:
C0:= [0.277164,0.340257,0.408617,0.581345,0.649268]:
plot(
     [seq([X(C0[1],m)(c), c, c= 0..1], m= M)],
     view= [0..1, 0..1], legend= M, labels= [x, C(x)]
);

@smith_alpha 

The 6 is the overall length, in characters, of the formatted number, not counting the "e" or the ".".

Note that my procedure MyPrint works without needing the number of digits or characters specified. It counts the digits itself.

@Kitonum Your example is not what the OP meant. All of the following steps work fine:

S:= Sum(w[k], k= 1..n);
P:= randpoly(x);
eval(P, x= S);
algsubs(S= Sw, %);
eval(%, Sw= S);

Please post the system of equations.

Having seen Kitonum's solution, I must say that my solution takes a risk of performing a mathematically invalid operation. If the expression had contained a stray 1/(T[1]*V[2]) (perhaps in one of the e exponents), my evalindets would've changed it to 1.

Nonetheless, I understand down to the smallest detail what evalindets does. I can't say the same for algsubs; it is somewhat unpredictable to me.

@Kitonum 

I admire your diligence in going through so many steps to post a neat expression to MaplePrimes. I don't think that I have the patience for that.

If you're going to be using the green arrow anyway, why not just upload a worksheet??

@nm 

Note that Maple stores a product-with-denominator P:= (a*b)/(c*d) as '`*`'(a, b, 1/c, 1/d), so a*b is a "subproduct" of P, but c*d is not. That is why your algsubs did not work. But 1/(c*d) is a subproduct, which is why Kitonum's worked.

I find algsubs unpredictable; the documentation is not clear. How do you know that algsubs(1/(a*b)= U, B) will not replace a*b with 1/U? I guess that you can't be sure until you see that there are no Us in the final expression.

@Kitonum 

How did you put your expression into your MaplePrimes post? Your expression is exactly the same as mine, which was cut-and-paste, yet yours is so much neater (and blue).

I asked you this before, but you did not answer: ARE THEY LINEAR EQUATIONS? Is some significant subset of them linear?

I'm warning you: People in this group generally do not take kindly to repeated Questions. Posting a followup Reply to your earlier Question puts it at the top of the Active Conversations list, so you do not gain any extra visibility by re-asking the Question. All it does is make angry the people who might answer your Question.

@kyleevanslee 

Applying my technique to a system with 11^2 variables is not at all feasible; there are far too many possibilities. However, if it could be done, then it'd be fairly easy to divide the solutions into equivalence classes in the way that you said.