multi variables in an array and solved by multi-equations

Hi All

I searched almost the entire community for the answer and did not find it. the original code:

syms xx yy zz 
eqn1=xx+yy==6;
eqn2=xx/yy==4;
eqn3=zz+yy==3;

AAA=[eqn1 eqn2 eqn3];

B=[xx yy zz];
B=vpasolve([AAA]);

The results is still:

EDU>> B

B = 

      xx: [1x1 sym]
      yy: [1x1 sym]
      zz: [1x1 sym]

but if I use the regular form:

[xx yy zz]=vpasolve([AAA]);

the system will give me the right answer.

The question is: How can I solve multi-variables in an array by multi-equations? this program is just a testing trial for another program. I need to find out how to solve the array instead of just typing all variables.

Thank you very much!

[Karan Gill]

The doc describes this in the 3rd example here. Is the example unclear?

Karan (Symbolic doc)

Answer 1

[Walter Roberson]

B=vpasolve([AAA]); is giving you the right answer. It is returning a struct with one field for every variable. If you need to extract individual variable, you can refer to them in the field, such as B.yy . If you need to subs() in all of the values into equations, you can subs() the structure, such as subs(eqns, B) which will be equivalent

fn = fieldnames(B);
vals = struct2cell(B);
temp = [fn(:).'; vals(:).'];
subs(eqns, temp{:})

that is, it would be equivalent to

   subs(eqns, 'xx', B.xx, 'yy', B.yy, 'zz', B.zz)

[shanghua chen]

Hi Walter

thank you for your response!

I changed the code according to your reply:

syms xx yy zz 
eqn1=xx+yy==6;
eqn2=xx/yy==4;
eqn3=zz+yy==3;

AAA=[eqn1 eqn2 eqn3];

B=[xx yy zz];
B=vpasolve([AAA]);

fn = fieldnames(B);
vals = struct2cell(B);
temp = [fn(:).'; vals(:).'];
subs(AAA, temp{:})

and it gave:

EDU>> subs(AAA, temp{:})
Error using sym.subs
Too many input arguments.

How do I solve it? Thank you!

[Walter Roberson]

syms xx yy zz 
eqn1=xx+yy==6;
eqn2=xx/yy==4;
eqn3=zz+yy==3;

AAA=[eqn1 eqn2 eqn3];
B = vpasolve(AAA);

subs(AAA, B)

The above is all that is needed.

Sorry, my previous code involving fieldnames was not valid. I should have said that

subs(AAA, B)

is equivalent to

fn = fieldnames(B);
vals = struct2cell(B);
subs(AAA, fn, vals)

in this case that would be equivalent to

subs(AAA, {'xx', 'yy', 'zz'}, {B.xx, B.yy, B.zz})

[shanghua chen]

Hi Walter

I really appreciated your prompt reply.

I tried again on Matlab:

syms xx yy zz 
eqn1=xx+yy==6;
eqn2=xx/yy==4;
eqn3=zz+yy==3;

AAA=[eqn1 eqn2 eqn3];

B=[xx yy zz];

B= vpasolve(AAA);

subs(AAA, B);

and it did not return exact solution of 'xx', 'yy' or 'zz'. if I type 'B' in the command window, it returned:

EDU>> B

B = 

      xx: [1x1 sym]
      yy: [1x1 sym]
      zz: [1x1 sym]

thank you!

[Walter Roberson]

The line

B=[xx yy zz];

is not doing anything useful there. You overwrite B on the next line, which is

B= vpasolve(AAA);

Also,

subs(AAA, B)

is not intended to return exact solutions of xx, yy, or zz: it is intended to use their values in further computation.

To view the values, you might be tempted to use

subs([xx yy zz], B)

However in the general case, the solutions in B might be vectors for each entry rather than scalars for each entry, and then it becomes difficult to read out which value corresponds to which variable. So you could do

   for fn = fieldnames(B).'; disp(fn{1}); disp(B.(fn{1})); end

[shanghua chen]

thank you! I will find out more.