Question 1202869
<pre>
x + 2y - z = 2
x + yz + zx = 3
2) x +yz = 2
   y + zx = 2
   z + xy = 2

I'll do # 1 for you. Apply the same concept and you should have no problems doing # 2.

       x + y + z = 3______x + y = 3 - z --- eq (i)
                     x + 2y - z = 2 ------- eq (ii)
                    x + yz + xz = 3 ------- eq (iii)

                   2x + 2y + 2z = 6 ----- Multiplying eq (i) by 2 ------ eq (iv)
                     x + 2y - z = 2 ----- eq (ii)
                         x + 3z = 4 ----- Subtracting eq (ii) from (iv) 
                              x = 4 - 3z ----- eq (v)

x + yz + xz = 3____x + (y + x)z = 3 --- eq (iii)
                   x + (3 - z)z = 3 --- Substituting 3 - z for x + y in eq (iii)
                     {{{matrix(1,3, x + 3z - z^2, "=", 3)}}} --- eq (vi)

                 {{{matrix(1,3, 4 - 3z + 3z - z^2, "=", 3)}}} ----- Substituting 4 - 3z for x in eq (vi)
                         {{{matrix(5,3, 4 - z^2, "=", 3, - z^2, "=", 3 - 4, - z^2, "=", - 1, z, "=", " "+- sqrt((- 1)/(- 1)), z, "=", ""+- 1)}}}  

                             <font color = blue><font size  = 4><b><u>z = 1</font></font></b></u>
                              x = 4 - 3z ------ Substituting 1 for z in eq (v)
                              x = 4 - 3(1)
                              x = 4
                              x = 1

                          x + y = 3 - z
                          1 + y = 3 - 1 ------- Substituting 1 for x, and 1 for z in eq (i)
                              y = 2
                              y = 1
                    <font color = blue><font size  = 4><b>(x, y, z) = (1, 1, 1)</font></font></b>

                             <font color = blue><font size  = 4><b><u>z = - 1</font></font></b></u>
                              x = 4 - 3z ------ eq (v)
                              x = 4 - 3(- 1) -- Substituting - 1 for z in eq (v)
                              x = 4 + 3
                              x = 7

                          x + y = 3 - z
                          7 + y = 3 - - 1 ----- Substituting 7 for x, and - 1 for z in eq (i)
                          7 + y = 4
                              y = 4 - 7
                              y = - 3
                    <font color = blue><font size  = 4><b>(x, y, z) = (7, - 3, - 1)</font></font></b>

The fact that x, y, and z are purported to have DIFFERENT values would suggest to me that
<font color = blue><font size  = 4><b>(x, y, z) = (1, 1, 1)</font></font></b> is NOT the solution set, but <font color = blue><font size  = 4><b>(x, y, z) = (7, - 3, - 1)</font></font></b> instead,
would be the CORRECT solution set!!

I don't think it has to be stated that x, y, and z are different numbers. The fact that they're 
given DIIFERENT variables would automatically mean that they are.

I believe, wholeheartedly, that a note should've been attached to this problem, stating that {{{x <> y<> z}}}.</pre>