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Two Equations, Two Unknowns


Date: 7/27/96 at 2:17:49
From: Anonymous
Subject: Two Equations, Two Unknowns

For the equations

          4    4      4
         x  + y  =   b  
  

         x  + y  =   a

solve for x and y in terms of a and b...

I've tried it this way:

1. from the second equation : a-x = y
2. subs into the first and I get the polynomial equation:

      4     3      2       3     4    4
    2x  - 4x  + 6ax  + - 4a x + a  - b  = 0

but I could not solve it... I know that there have to be 4 answers and 
the sum of them must be 2 and the product is 0.5 (a^4 - b^4) but then 
I get stuck. Please help me.


Date: 7/27/96 at 14:38:15
From: Doctor Pete
Subject: Re: Two Equations, Two Unknowns

Let
     R = a^4 + b^4 ,
     S = 3a^2 +- 2*Sqrt[2*R] .

(the "+-" symbol is the usual 'plus or minus' notation.)  Then

     {(x,y)} = {((a+Sqrt[-S])/2, (a-Sqrt[-S])/2)} .

Since the original system is symmetric in x and y, I have not written 
out the full solution set (which should be obvious).  Thus, we have a 
total of four possible solutions, two of which arise due to symmetry.  
This result was obtained through Mathematica, though I am fairly 
certain it can be solved by hand, looking at the simplicity of the 
solution.

To be honest, I haven't really thought about this problem; if you're 
more interested in how to obtain the solution rather than the result, 
then I might suggest trying to solve the above quartic in x by 
assuming that it has a factorization of the form 
(x^2 + px + q)(x^2 + rx + s) = 0, and then finding p, q, r, and s by 
expanding and equating coefficients.  If successful (I haven't worked 
it out yet), then you get two quadratics (which, looking from the 
solution, will most likely be related) which are easily solved using 
the quadratic formula.

Then again, it might not work.... ;)

Will post again if I figure it out.

-Doctor Pete,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   


Date: 7/28/96 at 0:55:37
From: Anonymous
Subject: Re: Two Equations, Two Unknowns

why do you let

> R = a^4 + b^4 ,
> S = 3a^2 +- 2*Sqrt[2*R] 

? I don't quite understand. And how do you solve for x and y from 
these two equations?  Thanks again!


Date: 7/28/96 at 17:7:15
From: Doctor Pete
Subject: Re: Two Equations, Two Unknowns

Well, last time I mailed you, I was only able to find the solution, 
not the *method*, using a computer program called Mathematica.  
So R and S are only labels to make writing the solution easier.  I'm 
very sorry for the confusion.

In any case, I have found a means of solving these equations.  From 
equation 2,

     x+y = a  ==> y = a-x

substituting into equation 1 gives

     x^4 + (a-x)^4 = b^4
 ==> 2x^4 - 4ax^3 + 6a^2 x^2 - 4a^3 x + a^4 - b^4 = 0
 ==>  x^4 - 2ax^3 + 3a^2 x^2 - 2a^3 x + (a^4 - b^4)/2 = 0 ,

as you've noted already.  Now, here's the nasty part.  We will factor 
the above 4th degree polynomial by assuming that it factors into two 
quadratics; that is,

     (x^2 + px + q)(x^2 + rx + s) = x^4 - 2ax^3 + 3a^2 x^2 - 2a^3 x
                                        + (a^4 - b^4)/2
                                  = 0 ,

and our goal then is to multiply out the two quadratics on the left 
and solve for the unknowns p, q, r, and s by equating the coefficients 
on the left with those on the right.  So

     (x^2 + px + q)(x^2 + rx +s) = x^4 + (p+r)x^3 + (pr+q+s)x^2 + (ps+
qr)x  + qs ,

and if we match up the coefficients, we see that

          p + r = -2a                      [equation 1]
     pr + q + s = 3a^2                     [equation 2]
        ps + qr = -2a^3                    [equation 3]
             qs = (a^4 - b^4)/2 .          [equation 4]

Now, at first this seems even more difficult than solving the original
problem, since there are more equations, but it's not.  Take the first 
two equations in this system;

          p + r = -2a
     ==>      r = -(2a+p) ,

     pr + q + s = 3a^2
 ==>         pr = 3a^2 - (q+s) .

Substituting the first into the second, we eliminate r and solve for 
p:

       -(2a+p)p = 3a^2 - (q+s)
   ==>  p^2 + 2ap + 3a^2 - (q+s) = 0

   ==>  p = (-2a +- Sqrt[4a^2 - 4(3a^2 - (q+s))])/2
          = -a +- Sqrt[a^2 - 3a^2 + q+s]
          = -a +- Sqrt[-2a^2 + q+s] .

This gives us two solutions.  Note that p and r must have the same 
solution set (similarly, q and s also have the same solution set), 
because the quadratics they represent are interchangeable.  This is 
important, because from this we know that either choice of p will 
work.  So we may let

     p = -a + Sqrt[-2a^2 + q+s] ,

and thus

     r = -a - Sqrt[-2a^2 + q+s] .

(Note this works, since p+r = -2a, which is equation 1.)  Now, take 
equation 3, and substitute:

     ps + qr = -2a^3
 ==> (-a + Sqrt[-2a^2 + q+s])s + (-a - Sqrt[-2a^2 + q+s])q = -2a^3
 ==> (-a)(q+s) + (s-q)Sqrt[-2a^2 + q+s] = -2a^3
 ==>             (s-q)Sqrt[-2a^2 + q+s] = -2a^3 + a(q+s)
                 (s-q)Sqrt[-2a^2 + q+s] = a(-2a^2 + q+s) .

Now, note that if q+s = 2a^2, both sides of this equation will be 
zero. Thus q+s = 2a^2 is a condition that solves equation 3.  Thus

     p = -a + Sqrt[-2a^2 + q+s] = -a + Sqrt[-2a^2 + 2a^2] = -a
     r = -a - Sqrt[-2a^2 + q+s] = -a - Sqrt[-2a^2 + 2a^2] = -a,

so p and r both equal -a.  Finally, equation 4 allows us to solve for 
q and s:

      qs = (a^4 - b^4)/2
     q+s = 2a^2
 ==>   s = 2a^2 - q

 ==> q(2a^2 - q) = (a^4 - b^4)/2
 ==> q^2 - 2a^2 q + (a^4 - b^4)/2 = 0

 ==> q = (2a^2 +- Sqrt[4a^4 - 4(a^4-b^4)/2])/2
       = a^2 +- Sqrt[a^4 - (a^4-b^4)/2]
       = a^2 +- Sqrt[(a^4+b^4)/2] .

Remembering that q and s must have the same solution sets (again due 
to symmetry), it follows that we may have

     q = a^2 + Sqrt[(a^4+b^4)/2] ,
     s = a^2 - Sqrt[(a^4+b^4)/2] ,

and thus our original polynomial has the factorization

     (x^2 - ax + q)(x^2 - ax + s) = 0 .

Thus we may solve for x by solving each quadratic individually; we 
have

     x = {(a +- Sqrt[a^2 - 4q])/2 , (a +- Sqrt[a^2 - 4s])/2}

         a +- Sqrt[-3a^2 +- 2*Sqrt[2(a^4+b^4)]]
       = -------------------------------------- ,
                           2

which is precisely what Mathematica found.  And because of symmetry, 
this solves y automatically as well.  So for one particular solution, 
choose x by picking the first plus/minus to be plus, and the second 
plus/minus to be minus; then y will be the same as above except you 
pick the first plus/minus to be *minus*, and the second to be *minus*.  
That is, you choose opposite signs on the outer plus/minus, and the *
same* sign on the inner one.  We denote this by saying

         a +- Sqrt[-3a^2 +- 2*Sqrt[2(a^4+b^4)]]
     x = -------------------------------------- ,
                           2

         a -+ Sqrt[-3a^2 +- 2*Sqrt[2(a^4+b^4)]]
     y = -------------------------------------- .
                           2

Note I have switched the order of the outer plus/minus symbol on y, so 
it reads "minus/plus."  This completes the solution of your system.  
(Whew!)

Now, I'd consider this quite difficult for a high-school level 
problem, primarily because you have to make several observations, one 
of which is assuming that the quartic you had found has a nice 
factorization.  The second difficult step is being able to solve for 
p, q, r, and s - this is the most computational part of the problem.

For a somewhat easier problem, can you solve

     x^3 + y^3 = b^3
     x   + y   = a    ?

(Hint:  factor x^3 + y^3.)

For a *very* nasty problem, do you think that there is a general 
method to solve

     x^n + y^n = b^n
     x   + y   = a

for positive integer n?

Hope I made sense - if I didn't, please don't hesitate to point out 
the unclear parts of my solution.

-Doctor Pete,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Basic Algebra
High School Linear Algebra

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