# Thread: Solving four equations for four variables ----crack it

1. ## Solving four equations for four variables ----crack it

Hi all,

I have four equations with four unknowns which cannot be said as neither polynomial nor algebraic. So I dont know how to solve with MATLAB.
The eqn are as follows:

3.52T-7145.3X^2+2.30237TX^2+19700Y^2-15.89TY^2+489+8.314T(ln(1-Y) – ln (1-X)) = 0 ....(1)

-60.19E-20T^7 – 11.274634T + 15.89TY^2 – 2.20171Z^2 + 3.0438TZ^3 – 19700Y^2 – 1583.4Z^3 + 6996.2Z^2 + 4183.124 + 8.314T(ln(1-Z) – ln (1-Y)) = 0 .........(2)

8.46T+ 19700 (1-X)^2 – 15.89T (1-X)^2 – 7145.3 (1-Y)^2 +2.30237(1-Y)^2 – 5510 + 8.314T (ln X – ln Y) = 0 ...........(3)

147.031E-20T^7-5.62777T-1.5219T(1-Z)^2 [1-2Z]+ 6204.5(1-Z)^2 – 0.67981 T(1-Z)^2 + 791.7(1-Z)^3 – 791.7{(Z)(1-Z)^2 } – 7145.3(1-Y)^2 +2.30237T(1-Y)^2 +1593.092 + 8.314T(ln Z – ln Y) = 0 .........................(4)

How to solve this as it is of mixed kind?
Pls help me to crack this.

Regards,
Surendar

2. 1: are you serious ?

2: why is there no TZ^2 term in equation(2) ?

3: have you tried setting T=0, thus (as example) equation(1) becoming:
7145.3X^2 - 19700Y^2 = 489

I guess you could rewrite each equation in terms of T, thus getting 4 equalities,
but only a deeply troubled mathematician would try that

3. ## Some corrections

These equations are solved equating chemical potentials in thermodynamics, so the data which are found in element data sheet are used to equate this and then finalised to these four equations. So i dont know the reason on what mathematical aspect u expect a TZ^2 term in eqn(2). Is there any reason or easy way of solving if it has that term?

Also, in third equation I missed "T" term in 2.30237T(1-Y)^2.

So the update third eqn is

8.46T+ 19700 (1-X)^2 – 15.89T (1-X)^2 – 7145.3 (1-Y)^2 +2.30237T(1-Y)^2 – 5510 + 8.314T (ln X – ln Y) = 0 ...........(3)

Sorry for the mistake.

Ok fine. If u say to make T as a function of X,Y and Z I think it is not so easy because T is not a single term and it is mixed with X,Y and Z as u can see in thse eqn.

4. Originally Posted by Surendra
So i dont know the reason on what mathematical aspect u expect a TZ^2 term in eqn(2).
NOTE: for your benefit, "u" is not a word; should be "you".

-60.19E-20T^7 – 11.274634T + 15.89TY^2 – 2.20171Z^2 + 3.0438TZ^3 – 19700Y^2 – 1583.4Z^3 + 6996.2Z^2 + 4183.124 + 8.314T(ln(1-Z) – ln (1-Y)) = 0 .........(2)

I noticed two Z^2 terms: 2.20171Z^2 and 6996.2Z^2, so assumed that
one of them should include a T, as this is the case wiyh Y^2 and TY^2.
So certainly "appears" that 2.20171Z^2 should be 2.20171TZ^2

5. Originally Posted by Surendra
... make T as a function of X,Y and Z I think it is not so easy because T is not a single term and it is mixed with X,Y and Z as u can see in thse eqn.
Seems easy enough; here's equation(1):
3.52T - 7145.3X^2 + 2.30237TX^2 + 19700Y^2 - 15.89TY^2 + 489 + 8.314T(ln(1-Y) – ln (1-X)) = 0 ....(1)
Get terms with T together:
3.52T + 2.30237TX^2 - 15.89TY^2 + 8.314T[ln(1-Y) – ln (1-X)] = 7145.3X^2 - 19700Y^2 - 489

T{3.52 + 2.30237X^2 - 15.89Y^2 + 8.314[ln(1-Y) – ln (1-X)]} = 7145.3X^2 - 19700Y^2 - 489
Isolate T:
T = (7145.3X^2 - 19700Y^2 - 489) / {3.52 + 2.30237X^2 - 15.89Y^2 + 8.314[ln(1-Y) – ln (1-X)]}

6. I suggest replacing all constants with letters that are not unknowns. It will make it alot less confusing. There is a generic method to solve $n$ equations of $n$ variables that involves rearranging each equation in a well-defined way, however if it is not possible to rearrange any of the equations then the method fails (and so does every other algebraic one). And since the equations look very twisted and include one transcendental equation (natural logarithm), I doubt it is possible to achieve successful rearrangement on all four equations.

Is this system routinely solved in chemistry (edit : thermodynamics my bad) or did you just stumble upon it ?

EDIT : if this is actually possible then ... my bad again but anyway the system is way too messy, even looking at it burns my eyes ... I might give it a try tomorrow if the question is still open.

7. ## Hi all

I came to know about a new optimisation technique from my friend. It is "Particle Swarn Optimisation" where I think my problem is more related to the system. Do anyone have better experience of working in this technique?

8. Originally Posted by Surendra
... where I think my problem is more related to the system.
What do you mean by that?

9. I mean that the my problem is solving these equations and the values obtained if back calculated in that equations it gives the value as zero. so this particle optimisation also follows the same method to find the solution for non-linear equations.