Two possible solutions:
x = 3;
idx = 1;
for c = -2:1:2;
for q = -2:1:2;
g = (c*x)+q*(c+x);
% g at this point is a scalar value, since in each iteration q and c are scalars.
% If you only want the output where g is in range:
if g >= 5 & g <= 10
disp([ 'at c=', num2str(c), 'at q=', num2str(q), ',g=', num2str(g)])
% You might also want to save these values
C(idx) = c;
Q(idx) = q;
G(idx) = g;
idx = idx+1;
end
end
end
There is also a possibly faster solution by not using for-loops
x = 3;
% the base array, as you use it for q and c
a = -2:2;
% now create two vectors whose combination is always unique, have a look at the arrays to confirm this.
c = repelem(a, numel(a));
q = repmat(a, [1, numel(a)]);
% compute all possible values for g
g = c*x + q.*(c+x);
% logical indices for all the elements of g that are in range
inRange = g>= 5 & g<=10;
G = g(inRange);
C = c(inRange);
Q = q(inRange);
