Lesson Solving systems of non-linear algebraic equations with symmetric functions of unknowns

Solving systems of non-linear algebraic equations with symmetric functions of unknowns

Problem 1

Preliminary notice. Since the problem is a bit higher than the traditional school math, the solution is, correspondingly, 
a bit higher (but still understandable). 

We are given the numerical values for the functions x+y+z,  xy+yz+zx  and  . 
A remarkable fact is that the function  xyz  can be explicitly expressed via  x+y+z,  xy+yz+zx and . (I will do it later).

So, knowing the numerical values of  x+y+z,  xy+yz+zx  and , we can find the value of xyz.

Next, knowing the numerical values of  x+y+z,  xy+yz+zx  and  xyz, we can consider the polynomial of the variable "u"

P(u) = .   (2)

This polynomial is nothing else as 

P(u) = (u-x)*(u-y)*(u-z).     (3)

In other words, the polynomial (2) is factorable into (3).

Now, instead of solving the system (1), we can solve the polynomial equation 

P(u) = 0      (4)

for u. If we solve it (and when we solve it), its roots u = x, u = y and u = z will be the solution of the system (1).

Why this way is better than solving (1) directly?
Well, for example, you can (try to) solve the polynomial equation (4) graphically. 
Or apply other methods specific for polynomial equations. 
You will see it later.

Now I will implement this methodology.

We will do it step by step:

1.   = .


2.    = 
   = x(xy+xz) + y(xy+yz) + z(xz+yz) = 
   = x(xy+xz+yz-yz) + y(xy+yz+xz-xz) + z(xz+yz+xy-xy)
   = x(xy+xz+yz) + y(xy+xz+yz) + z(xy+xz+yz) - x(yz) - y(xz) - z(xy)
   = (x+y+z)(xy+xz+yz) - 3xyz.


3.  Therefore, 

      = ,   or

      = ,   or

      = ,   or

     xyz = . 


4.   Thus

     xyz =  =  = 48.

So, our polynomial (2) is

    P(u) = ,

and we need solve this polynomial equation (4)

     = .

First, let's do it graphically.

Figure. Plot P(u) =

Do you see the roots? But of course, they are u=2, u=4 and u=6.

And you can check it manually substituting these values into the polynomial.

Or you can apply the rational roots theorem.

According to this theorem, all the roots are among the integer divisors of the number 48, and you have only finite number 
of options to check. It is your other method to find the roots.

Problem 2

  x + 2y  + 4z = 9, 
4yz + 2xz + xy = 13,
      xyz      =  3.

Introduce new variables 

p = x,  q = 2y  and  r = 4z.           (1)

Then  pq = 2xy,  pr = 4xz,  and  qr = 8yz,  so  pq + pr + qr = 2xy + 4xz + 8yz = 2*(xy + 2xz + 4yz).


Therefore, the original system takes the form

p + q + r = 9,          (2)    

pq + pr + qr = 26,      (3)    (26 = 2*13)

pqr = 24.               (4)    (24 = 8*3)


Thus, according to the Vieta's theorem,  p, q and r are the roots of this polynomial equation of the degree 3:

t^3 - 9t^2 + 26t  - 24 = 0.


By applying the "Rational root theorem", you can find the integer roots of this equation.

If exist, they are among the integer divisors of the constant term 24, i.e. among the number set

{+/-1, +/-2, +/-3, +/-6, +/-12, +/-24}.


Or, you can plot a graph, and it will tell you/ (will show you) what the roots are:





Plot y = 


and the plot says that the roots are t = 2, 3 and 4.


Thus the solution (p,q,r) to the system  (2), (3), (4)  is ANY permutation of numbers  (2,3,4).
Any of 3! = 1*2*3 = 6 possible permutations.

Now we must return from variables p, q and r to the original variables x, y and z using formulas (1).

Since the original system is EQUIVALENT to the system (2), (3), (4) due to substitution (1),

it implies that the following values (triples) of x, y and z are the solutions to the original system:

    1)  (p,q,r) = (2,3,4)  ====> x= 2,  y= 3/2,  z=  1;

    2)  (p,q,r) = (4,2,3)  ====> x= 4,  y= 1,    z= 3/4;

    3)  (p,q,r) = (3,4,2)  ====> x= 3,  y= 2,    z= 1/2;
 
    4)  (p,q,r) = (3,2,4)  ====> x= 3,  y= 1,    z=  1;  

    5)  (p,q,r) = (4,3,2)  ====> x= 4,  y= 3/2,  z= 1/2;

    6)  (p,q,r) = (2,4,3)  ====> x= 2,  y= 2,    z= 3/4.


You can check it manually (as I did it . . . ).


Answer.  The given system has 6 solutions listed above.

Problem 3

Let me introduce new variables u = , v = ,  and w = .

Then I can rewrite the given equations in this form:

u + v + w = 10,   (1)

 = 38,    (2)

u*v*w = 30.       (3)


From (1) and (2)

 = 100 =  =  = 38 + 2(uv + uw + vw),

hence,  2(uv + uw + vw) = 100 - 38 = 62,   and  uv + uw + vw = 31.


Now I can rewrite the system (1),(2),(3) in the form

u + v + w = 10,       (1')

uv + uw + vw = 31,    (2')

u*v*w = 30.           (3')


It means ( !! The Vieta's formulas !! ) that u, v ans w are the roots of this cubical equation

X^3 - 10X^2 + 31X - 30 = 0.   (4)


Now you can apply the "Rational roots Theorem" which says that the rational (in particular, integer) roots of the equation (4) 
are among the divisors of the constant term 30.

Or even better, you can make a plot of the left side polynomial which (the plot) will say you a lot about the roots:





Plot y = 


From the plot, it is clearly seen that the roots u, v and w are 2, 3 and 5 (from smaller to greater).


You can check it manually.


Hence, the solution to the original system is (x,y,z) = (4,9,25) and all possible permutations.