How would I write a code to solve a system of equations?

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Omar 2013-10-19

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回答： anjali wavhale 2020-12-1

I need to solve a system of equations but I do not know the syntax for doing so. The equations are:

q1 = (1.53*10^-5)*t1^4 + (1.67*10^-2)*t1^3 + 6.85*t1^2 + 2746*t1 - 57793

q2 = 13.3*(t1-t2)

q3 = (1.53*10^-5)*t2^4 + (1.67*10^-2)*t2^3 + 6.85*t2^2 + 4846*t2 - 49161

also; q1=q2=q3

I'm new to MATLAB and not sure how to handle this many variables along with the scientific notation.

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Qasim Manzoor 2013-10-19

do you have to use a mathematical method or any way would be fine?

Omar 2013-10-20

I don't know what you mean exactly. How would you solve this in a non-mathematical way?

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sixwwwwww 2013-10-20

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编辑：sixwwwwww 2013-10-20

在 MATLAB Online 中打开

Dear Omar, you can solve your system of equations using the following way:

t = sym('t%d', [1 2]);
q1 = 1.53e-5 * t(1)^4 + 1.67e-2 * t(1)^3 + 6.85 * t(1)^2 + 2746 * t(1) - 57793;
q2 = 13.3 * (t(1) - t(2));
q3 = 1.53e-5 * t(2)^4 + 1.67e-2 * t(2)^3 + 6.85 * t(2)^2 + 4846 * t(2) - 49161;
[solutions_t1, solutions_t2] = solve(q1 == q2 == q3, t(1), t(2))

[solutions_t1, solutions_t2] = solve(q1 == q2, q2 == q3, t(1), t(2))

You can check which solutions are better because in first case you will get 4 solutions for both t1 and t2 and in second case you will get 16 solutions for both t1 and t2. I hope it helps. Good luck!

When you put q1 = q2 = q3 and then you solve then you will get one equation in which you will have two variables t1 and t2. So for any value of t1 you will get corresponding value for t2. for example for this code line:

[solutions_t1, solutions_t2] = solve(q1 == q2 == q3, t(1), t(2))

I get the following solution and MATLAB uses the mathematical approach which i just described above:

solutions_t1 =
                                           20.086842601695534071659582222863
                                           -818.0065723267874137510207732586
   - 146.79176912438219320848725608014 + 456.44240227248617435061388081659*i
   - 146.79176912438219320848725608014 - 456.44240227248617435061388081659*i
solutions_t2 =
   0
   0
   0
   0

As you can see here MATLAB assumes value 0 for t2(four 0s because of polynomial of degree 4) and computes the values for t1 because here you have just one equation for t1 and t2. So if you will assume different set of values for t2 then you will get different set of values for t1 and in this way there will be infinite solutions to this problem.

Omar 2013-10-20

ok I got what I'm looking for. Thank you for your help.

sixwwwwww 2013-10-20

You are welcome dear

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Walter Roberson 2013-10-20

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在 MATLAB Online 中打开

{q = 2003.839045, t1 = -819.5264017, t2 = -970.1909915},
{q = -10982.94224, t1 = -817.9943073, t2 = 7.790823431},
{q = 133.2164095, t1 = 20.04280634, t2 = 10.02653495},
{q = 13235.45859, t1 = 24.30348638, t2 = -970.8437757}

provided that you only want real-valued solutions.

This was done by substituting t1 for t(1) and t2 for t(2), and then plugging the three q equations into solve()

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David 2013-10-21

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编辑：David 2013-10-21

在 MATLAB Online 中打开

In addition to the answers already posted, be sure to put sym() around each of your equations. That is to say, do the following:

syms t1 t2
 q1 = sym( 1.53e-5*t(1)^4 + 1.67e-2*t(1)^3 + 6.85*t(1)^2 + 2746*t(1) - 57793 );
 q2 = sym( 13.3*(t(1) - t(2)) );
 q3 = sym( 1.53e-5*t(2)^4 + 1.67e-2*t(2)^3 + 6.85*t(2)^2 + 4846*t(2) - 49161 );
[solutions_t1, solutions_t2] = solve(q1 == q2, q2 == q3, t1, t2);
solve requires that your equations be symbolic objects which is guaranteed by using sym().

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Walter Roberson 2013-10-21

在 MATLAB Online 中打开

As long as anything in the equation is sym, the whole expression will be treated as sym. So

t = sym('t%d', [1 2]);

is enough to make the equations sufficiently sym() without requiring sym() around the equations.

David 2013-10-22

Oh? This seemed to correct a problem I was having earlier in using the solve function. You actually saw that question and made useful suggestions, though suggestions that did not quite do the trick. Maybe we are both missing something or maybe it's just me...

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