Question 696814
Given:
(x+y)(x+y+z)=66----------------------------------------eq.1
(z+y)(x+y+z)=99----------------------------------------eq.2
(z+x)(x+y+z)=77----------------------------------------eq.3

Solution : Add all 3 equations
{{{(x+y)(x+y+z)+(x+y)(x+y+z)+(z+x)(x+y+z)=242}}}
{{{(x+y+z)2(x+y+z) = 242}}}
{{{(x+y+z)^2 = 121 = 11^2}}}
x+y+z = 11  -------------------------------------------eq.4

substitute eq.4 in eq. 1 , eq.2 and eq.3 , we get

x+y  = 6------------------------------------------------eq.5
y+z  = 9------------------------------------------------eq.6
x+z  = 7------------------------------------------------eq.7

Add  eq.5 and 6, we get

{{{x+2y+z = 15}}}---------------------------------------eq.8

Subtract eq.4 from eq.8, we get

y = 4 --------------------------------------------------a

Put a in eq.5 and eq.6, we get,

X = 2 , Z = 5, Y = 4