math olympiad question.. try ur best to solve!?

what are all ordered triples of real numbers (x,y,z) that satisfy

(x+y)(x+y+z) =120

(y+z)(x+y+z)=96

(x+z)(x+y+z)=72

show the work.. if u get this question rite.. u r a math geniuss!

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Top 1 Answers

A few days ago

shmnih

Favorite Answer

(x+y)(x+y+z) is

x^2 + y^2 + 2xy + xz + yz

(y+z)(x+y+z) is

y^2 + z^2 + 2yz + xz + yz

(x+z)(x+y+z) is

x^2 + z^2 + 2xz + xy + zy

adding all three you get

2x^2 + 2y^2 + 2z^2 + 4xy + 4yz + 4zx

and simplyfying this you get

2(x+y+z)^2 = 120 + 96+ 72 = 288

(x+y+z)^2 = 144

or x+y+z = + or – 12

substitute this in each equation and when i used +12

I got x = 4, y = 6 and z=2

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