Question

Three variables x,y,z exist such that x^2+y^2+z^2 = 50 , xy + yz + zx = 47 ; xyz = 60

(i) (x+y+z)^3 = ? -- {i know this one!}
(ii) What's the value of x ?
(iii) What's the value of y ?

Answer 1

This is a system of equations. First, you need to find the value of x, y, and z. I recommend doing this by using the last equation:

xyz = 60

x = 60/yz

Then you can substitute that in and find the value of y, z, and x.

Answer 2

@thechocoluver445 : I did the very same thing.. but at the end, i got an equation, that i wasnt able to solve.. and that ruined my whole work! :/

Answer 3

(x+y+z)^2= x^2+y^2+z^2+2(xy + yz + zx)=50+2*47=154
so we have x+y+z=sqrt(154) or x+y=sqrt(154)-z (I)

xyz=60 so xy=60/z (II)

xy + yz + zx=47
from (I) and (II)
xy + yz + zx=xy+z(x+y)=60/z+z(sqrt(154)-z)=47

then u may solve \[-z^3+\sqrt{154} \ z^2-47z+60=0\]

Answer 4

for solving such equations
http://www.numberempire.com/equationsolver.php

Answer 5

@mukushla : Dear, you probably did the steps correctly , but x+y+z = 12.. as in that first answer what you got as 154 , is actually 144 ..

Answer 6

oh much better

Answer 7

(x+y+z)^2= x^2+y^2+z^2+2(xy + yz + zx)=50+2*47=144
so we have x+y+z=12 or x+y=12-z (I)

xyz=60 so xy=60/z (II)

xy + yz + zx=47
from (I) and (II)
xy + yz + zx=xy+z(x+y)=60/z+z(12-z)=47

we must solve

\[−z^3+12z^2−47z+60=0\]

Answer 8

thats very interesting
the solutions for z are
z=3
z=4
z=5

Answer 9

(x,y,z)=(3,4,5)

Answer 10

Exactly .. !! Thanx to that "equation solver" of yours.. Helped me a ton! :))