the number of real solutions of the systemx+y+z=1,2xy-z2=1 isI am getting 1 as the ans but answer given is 3

by simple analysis we see that 

x=1 y=1 z=-1 satisfy both the equations 

z= 1- ( x+y)

put in second equation 

2xy - { 1-( x+y)}^2  =1 

2xy -1 - (x+y)^2 + 2(x+y)=1

2(X+Y)-(X^2 +Y^2)=2 

only has one solution