How to solve a multi variable inequality, where variables are a given set of numbers?

syms B D L;
A=2860;
k=5900:20:7200;
B=k(randperm(numel(k),1));
C=34620;
l=27400:20:28240;
D=l(randperm(numel(l),1));
E=C-D;
F=E-A;
Q=5.9;
G=F/Q;
H=B/G;
m=4.5:0.1:5.3;
L=m(randperm(numel(m),1));
M=(H/(L+100))*100;
S=2.160;
N=(M/S)*100;
P=98.00<=N<=99.99;
W=solve(P,[B D L])

Value of B should be within 5900:20:7200. Value of D should be within 27400:20:28240. Value of L should be within 4.5:0.1:5.3. Value of N should be within 98.00:0.01:99.99.

I want to get multiple combinations of solutions in tabular form for [B D E F G H L M N].

Answer 1

[Dayalram Dayalram]

After doing little research I found out that it is not a problem of algebra, actually it is a problem of simple for loop problem. Here is how I solved it:

function compaction_values=fdd(lower,higher)
%FDD Field dry density.
%    FDD(lower,higher) prints all combinations of values of A,B,C,D,E,
%    F,G,H,L,M and Field Compaction accordingly, where field compaction's
%    highest value is equal to input argument 'higher' and field compaction's
%    lowest value is equal to input argument 'lower'.
%    
%    Note: The values of inputs 'lower' and 'higher' should be nonnegative
%    real numbers. Negative values will print an error message.
if nargin ~= 2
    error('Number of inputs must be two.');
elseif ~isscalar(lower) || ~isscalar(higher) || lower<0 || higher<0 || isempty(lower) || isempty(higher)
    error('Inputs must be nonnegative scalars.');
else
    A = 2860;
    C = 34620;
    S = 2.160;
    Q = 1.349;
    count = 0;
    compaction_values=[];
    for B = 6000:20:7100
        for D = 27440:20:28180
            for L = 5:-0.1:4.5
                E = C-D;
                F = E-A;
                G = F/Q;
                H = B/G;
                M = (H/(L+100))*100;
                compaction = (M/S)*100;
                if compaction >=lower && compaction <= higher %This statement forces the function to print only needed values.
                   fprintf('%.0f  %.0f  %.0f  %.0f  %.0f  %.0f  %.0f  %.2f  %.1f  %.2f  %.2f\n',A,B,C,D,E,F,G,H,L,M,compaction);
                   compaction_values=[compaction_values;A,B,C,D,E,F,G,H,L,M,compaction];
                   count = count+1; %Counts the number of combinations.
                end
            end
        end
    end
    fprintf('\ncount=%d\n',count);
end
end

You might have noticed some chages in data from the actual question but it's the same type of question.