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Solutions of a System (Posted on 2009-10-26) Difficulty: 3 of 5
In the following system of equations x,y,z are variables and k is a constant. All values are real numbers.
  • x + y*z = k
  • y + x*z = k
  • z + x*y = k
Determine the relationship between k and the number of the solutions of the system.

  Submitted by Brian Smith    
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Solution: (Hide)
First, solve the system in terms of k
x + y*z = k
y + x*z = k
z + x*y = k

Take the difference of the first two equations
(x + y*z) - (y + x*z) = k - k
(x-y) + z*(y-x) = 0
(x-y)*(1-z)=0
x=y OR z=1
            case x=y                           OR       case z=1
-----------------------------------------------||--------------------------
Substitute x=y into the original system        ||Substitute z=1 in the
x + x*z = k  AND  z + x*x = k                  ||original system
                                               ||
Take the difference of these equations         || x + y = k
(x-z) + x*(z-x) = 0                            || x * y = k-1
x=z OR x=1                                     ||
     subcase x=z      OR     subcase x=1       ||Then (x,y,z) = (k-1,1,1)
-----------------------|-----------------------||or (x,y,z) = (1,k-1,1)
x,y,z are all equal    |x and y equal 1        ||
solve x^2+x = k        |then (x,y,z)=(1,1,k-1) ||
x=(-1+/-sqrt[4k+1])/2  |                       ||

There are five possible solutions total:
(x,y,z)=(  1,  1,k-1)
(x,y,z)=(  1,k-1,  1)
(x,y,z)=(k-1,  1,  1)
x=y=z = (-1 + sqrt[4k+1])/2
x=y=z = (-1 - sqrt[4k+1])/2

If k=2, there are two solutions (1,1,1) or (-2,-2,-2).
If k>-1/4 and k!=2, there are five solutions.
If k=-1/4, there are four solutions
If k<-1/4, there are three real solutions (and two complex solutions).
				

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionSpoiler (no proof)Steve Herman2009-10-26 14:11:40
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