If #xy+yz+zx=12-x^2=15-y^2=20-z^2# then #x+y+z=?#

2 Answers
Narad T.
Jul 29, 2018

The answer is #=+-6#

Explanation:

Firstly,

#xy+yz+zx=12-x^2#

#=>#, #xy+x^2+yz+zx=12#

#=>#, #x(x+y)+z(x+y)=12#

#=>#, #(x+y)(x+z)=12#

Therefore,

#{(x+y=+-3),(x+z=+-4):}#

Similarly,

#xy+yz+zx=15-y^2#

#xy+zx+y^2+yz=15#

#(x+y)(y+z)=15#

Therefore,

#{(x+y=+-3),(y+z=+-5):}#

And finally,

#xy+yz+zx=20-z^2#

#xy+zx+yz+z^2=20#

#(y+z)(x+z)=20#

Therefore,

#{(y+z=+-5),(x+z=+-4):}#

So,

#x+y+x+z+y+z=(+-3)+(+-4)+(+-5)=+-12#

#2(x+y+z)=+-12#

#x+y+z=+-12/2=+-6#

Before, I considered that #(x, y, z) in NN^3# but #(x,y,y) in ZZ^3# is also valid.

P dilip_k
Jul 29, 2018
  • #xy+yz+zx=12-x^2#

#=>xy+yz+zx+x^2=12#

#=>y(x+z)+x(z+x)=12#

#=>(x+y)(z+x)=3xx4.....[1]#

  • #xy+yz+zx=15-y^2#

#=>xy+zx+yz+y^2=15#

#=>x(y+z)+y(z+y)=15#

#=>(x+y)(y+z)=3xx5....[2]#

  • #xy+yz+zx=20-z^2#

#=>y(x+z)+z(x+z)=20#

#=>(y+z)(x+z)=5xx4....[3]#

It is obvious from [1], [2] and [3]

#((x+y)(y+z)(z+x))^2=3^2xx4^2xx5^2#

#=>(x+y)(y+z)(z+x)=pm(3xx4xx5).....[4]#

By [1] and [4]
#y+z=pm5#

By [2] and [4]
#z+x=pm4#

By [3] and [4]
#x+y=pm3#

Summing up we get

#2(x+y+z)=pm12#

#=>x+y+z=pm6#

Related questions
Impact of this question
2987 views around the world
You can reuse this answer
Creative Commons License