In Maxima, how can I solve this equation?

Question

In Maxima, how can I solve this equation? And actually I have 24 unknown value with lots of this kind of equation. How I can use Maxima to solve?

to know the unknown value: a, b, c, d

eq:[a*c=8,a*d=10,b*c=12,b*d=15,a+b=5,c+d=9];
solve(eq,[a,b,c,d]);

Answer 1

Your equations are, if I'm not mistaken, polynomials in several variables of degree 2 or greater (where the degree is equal to the maximum number of variables multiplied together). You say that you have more equations than unknowns.

If I'm not mistaken, in general there are no exact solutions, but you can look for a best-fitting solution via least squares. I don't know how many locally best-fitting solutions there might be. You might ask on math.stackexchange.com about the number of solutions and other characteristics of such equations.

Here is a way to look for a locally best-fitting solution in Maxima. Form a "figure of merit" expression (fom below) from the equations, and look for a local minimum of it. I'll use lbfgs to search for a numerical approximation.

(%i3) eq:[a*c=8,a*d=10,b*c=12,b*d=15,a+b=5,c+d=9];
(%o3) [a*c = 8,a*d = 10,b*c = 12,b*d = 15,b+a = 5,d+c = 9]
(%i4) eq0 : map (lambda ([e], lhs(e) - rhs(e)), eq);
(%o4) [a*c-8,a*d-10,b*c-12,b*d-15,b+a-5,d+c-9]
(%i5) eq2 : expand (eq0^2);
(%o5) [a^2*c^2-16*a*c+64,a^2*d^2-20*a*d+100,b^2*c^2-24*b*c+144,
       b^2*d^2-30*b*d+225,b^2+2*a*b-10*b+a^2-10*a+25,
       d^2+2*c*d-18*d+c^2-18*c+81]
(%i6) load (lbfgs);
(%o6) "/usr/local/share/maxima/branch_5_40_base_95_g4c9cbbd/share/lbfgs/lbfgs.mac"

(%i9) fom : apply ("+", eq2);
(%o9) b^2*d^2+a^2*d^2+d^2+2*c*d-30*b*d-20*a*d-18*d+b^2*c^2
             +a^2*c^2+c^2-24*b*c-16*a*c-18*c+b^2+2*a*b-10*b+a^2
             -10*a+639

Here fom is just the sum of squares of error (i.e. the discrepancy between the right hand side and left hand sides of the equations, squared and added up).

(%i10) lbfgs (fom, [a, b, c, d], [1, 1, 1, 1]/4, 1e-4, [1, 0]);
*************************************************
  N=    4   NUMBER OF CORRECTIONS=25
       INITIAL VALUES
 F=  6.198906250000000D+02   GNORM=  4.916696680699350D+01
*************************************************

   I  NFN     FUNC                    GNORM                   STEPLENGTH

   1    5     1.029984221465989D+01   3.303822830240210D+01   1.342303790326981D-01  
   2    7     1.858590097968602D+00   7.066597977731433D+00   2.567282877401947D-01  
   3    9     1.707935938873334D+00   7.250276888460332D+00   1.939008057328834D-01  
   4   10     1.071933775239074D+00   3.590373568045903D+00   1.000000000000000D+00  
   5   11     6.984083995048138D-01   2.736311274794537D+00   1.000000000000000D+00  
   6   12     2.756317886006059D-01   2.992718120186393D+00   1.000000000000000D+00  
   7   13     4.947211249555039D-02   2.105297770017910D+00   1.000000000000000D+00  
   8   14     8.047123928918154D-04   2.073845793942183D-01   1.000000000000000D+00  
   9   15     7.401148110375289D-06   2.195111965609954D-02   1.000000000000000D+00  
  10   16     3.129228502984915D-07   4.565302248467960D-03   1.000000000000000D+00  
  11   17     1.750777300912887D-09   4.710495250075657D-04   1.000000000000000D+00  

 THE MINIMIZATION TERMINATED WITHOUT DETECTING ERRORS.
 IFLAG = 0
(%o10) [a = 2.000003188592,b = 2.999996386563707,
        c = 4.000008973580663,d = 5.00000040529949]

I can guess that the a = 2, b = 3, c = 4, d = 5 might be an exact solution. I can verify that:

(%i11) subst ([a = 2, b = 3, c = 4, d = 5], fom);
(%o11) 0

That's encouraging in this case, but in general I think there won't be an exact solution near the approximate one. The approximate solution may be the best you can do.

Answer 2

With four unknowns you have to have four equations. Ignore one set of equations and run:

sol1: solve([a*c=8,a*d=10,b*c=12,a+b=5],[a,b,c,d]);
              [[a = 2, b = 3, c = 4, d = 5]]

Then, ignore an other set of equations and run again:

sol2: solve([a*c=8,a*d=10,b*c=12,c+d=9],[a,b,c,d]);
              [[a = 2, b = 3, c = 4, d = 5]]

If the two results are the same, then your system is consistent and has solution sol1 ≡ sol2.