Using Trial Solutions

Date: 07/28/99 at 15:12:15
From: Edward
Subject: Matrix and Multiple Unknowns

How do you do this?

     x - sqrt(yz) =  42
     y - sqrt(xz) =   6
     z - sqrt(xy) = -30

It seems easy but it is really HARD!

I think that it's kind of like this:

          x - sqrt(yz) + y - sqrt(zx) = 48   (add eq'n. 1 and 2)
   x + y - sqrt(z)[sqrt(y) + sqrt(z)] = 48   (factor sqrt(z))
     sqrt(z) = (48-x-y)/-[sqrt(y)+sqrt(z)]

Now, should I substitute my new equation into equation 1 or 2, OR 
should I square the equation and then substitute into equation 3?

Date: 07/28/99 at 18:13:50
From: Doctor Anthony
Subject: Re: Matrix and Multiple Unknowns

Write the equations in the form

     x-42 = sqrt(yz)   ->    (x-42)^2 = yz

      y-6 = sqrt(xz)   ->     (y-6)^2 = xz

     z+30 = sqrt(xy)   ->    (z+30)^2 = xy

Trial solutions are probably the quickest method for solving these 
equations. Looking at the equations it is probable that all three of 
x, y, z are multiples of 6. From third equation one at least of x or y 
will be greater than 30 and z is probably small. From the first 
equation x is greater than 42. If we try z = 6 we obtain

     (x-42)^2 = 6y     (y-6)^2 = 6x     36^2 = xy

and multiplying the first two of these equations

     (x-42)^2 (y-6)^2 =  36 xy

taking square roots of both sides

     (x-42)(y-6) = 6 (36) =  216

      xy - 42y - 6x + 252 = 216

     1296 - 42y - 6x + 36 = 0

                 6x + 42y = 1332

                   x + 7y = 222   and   xy = 1296

so                      x = 1296/y

              1296/y + 7y = 222

              1296 + 7y^2 = 222y

      7y^2 - 222y + 1296 = 0


         222 +- sqrt(222^2 - 36288)   222 +- 114
     y = -------------------------- = ----------
                     14                   14

       = 24   or   7.71428

So take y = 24

Then  x = 1296/24 = 54

And so the three roots are  x = 54,  y = 24,  z = 6

If you check these values in any of the original equations you will 
find that they satisfy those equations.

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/