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2014-12-22T20:19:38+05:30
Value of x is impossible with the given content.  x^{2} + y^{2} +   z^{2} =81
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2014-12-23T05:56:25+05:30

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you have given two equations only and there are three variables.  Hence one cannot solve for x as a numerical value.  But x can be found in terms of y and z.

     x + y + z = 15          --- 4
     (x + z) y + z x = 72
       (15 - y) y + z x = 72

          x = ( y^2 - 15 y + 72 ) / z  =  15 - y - z            ---  3

       then  y and z satisfy :  y^2 + z^2 - 15 (y+z) + y z +72 = 0      - -- 2

                   by solving this equation of 2nd degree,  y can be found in terms of z.

 Further,
        (x+y+z)^2 - 2 (xy+yz+zx) = 15^2 - 2 * 72
                x^2 + y^2 + z^2 = 81
                   81 - x^2  - 15 (y + z) + y z + 72 = 0

                    x^2 =  153 - 15 ( y + z ) + y z      --- 1
              
         substituting value of y found above,  x can be found in terms of z.

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