+1839 The system of equations
{xy}/{x + y} = 1,{xz}/{x + z} = 1, {yz}/{y + z} = 3
has exactly one solution. What is z in this solution?
+129852 (xy) / (x + y) = 1
(xz) /(x + z) =1
(yz) ( y + z) = 3
xy = x + y
xz = x + z → xz - x = z → x ( z - 1) = z → x = z / (z -1)
yz = 3(y + z) → yz = 3y + 3z → yz - 3y = 3z → y ( z - 3) = 3z → y = 3z / ( z -3)
xy = x + y
[ z / ( z -1) ] [3z / ( z -3) ] = z / (z - 1) + 3z / ( z -3)
[ 3z^2 ] / [ (z -1) ( z - 3) ] = z (z -3) + 3z ( z -1) / [ (z -3) (z -1) ]
3z^2 = z ( z -3) +3z ( z -1)
3z^2 = z^2 - 3z + 3z^2 - 3z
z^2 - 6z = 0
z ( z - 6 ] = 0
z = 0 (reject)
z = 6
y = 6
x = 6/5
