My second favourite problem ever

Taking the square root of each equation, $x+y = \pm 4 , y+z = \pm 6, x+z = \pm 9$ If you add all the 3 equations and divide by 2, you'll notice that there are $2^3$ possible values, of which 6 less than 3. The 3 roots which are greater than three are $\frac{11}{2} , \frac {19}{2} and \frac{7}{2}$

A total of 8 solutions are possible in this case.

i.) x+y =4; y+z = 6; x+z=9;

ii.) x+y =-4; y+z = 6; x+z=9;

iii.) x+y =4; y+z = -6; x+z=9;

iv.) x+y =4; y+z = 6; x+z=-9;

v.) x+y =-4; y+z = -6; x+z=9;

vi.) x+y =4; y+z = -6; x+z=-9;

vii.) x+y =-4; y+z = 6; x+z=-9;

viii.) x+y =-4; y+z = -6; x+z=-9;

Out of these 8 cases, only the first 3 have the required solution (x+y+z>3)