Problem:
If x,y, and z are positive numbers satisfying
x+1/y=4,y+1/z=1, and z+1/x=7/3
then what is the value of xyz=?
Answer Choices:
A. 2/3
B. 1
C. 4/3
D. 2
E. 7/3
Solution:
Note that (x+1/y)+(y+1/z)+(z+1/x)=4+1+7/3=22/3 and that
28/3=4⋅1⋅7/3=(x+1/y)(y+1/z)(z+1/x)=xyz+x+y+z+1/x+1/y+1/z+1/(xyz)=xyz+22/3+1/(xyz)
It follows that xyz+1/(xyz)=2 and (xyz−1)2=0. Hence xyz=1.
{OR}
By substitution,
4=x+y1=x+1−1/z1=x+1−3x/(7x−3)1=x+4x−37x+3
Thus 4(4x−3)=x(4x−3)+7x−3, which simplifies to (2x−3)2=0. Accordingly, x=3/2,z=7/3−2/3=5/3, and y=1−3/5=2/5, so xyz=(3/2)(2/5)(5/3)=1.
The problems on this page are the property of the MAA's American Mathematics Competitions