SOLUTION: Solve for x,y,z if (x+y)(y+z)=18,(y+z)(z+x)=30,(x+y)(x+z)=15

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Question 974920: Solve for x,y,z if (x+y)(y+z)=18,(y+z)(z+x)=30,(x+y)(x+z)=15
Answer by JoelSchwartz(130) About Me  (Show Source):
You can put this solution on YOUR website!
(x+y)=18/(y+z)
x+y=15/(x+z)
18/(y+z)=15/(x+z)
(18(x+z))/(y+z)=15
18(x+z)=15(y+z)
(y+z)=30/(z+x)
18(x+z)=15*(30/(z+x))
18(z+x)^2=15*30
(z+x)^2=25
z+x=+-5
5(y+z)=30
y+z=6
5(x+y)=15
x+y=3
x=3-y
z+3-y=5
z-y=2
z=2+y
y+2+y=6
2y=4
y=2
z=4
x=1