Absolute Value

How many different real-valued pairs (x,y) satisfy this system of two equations?
|x-y|=1
x/y=xy
Source: NCTM Mathematics Teacher, November 2006

SOLUTION
Suppose x-y>0
x-y=1\qquad\qquad (1)
x/y=xy\qquad\qquad\, (2)
y=x-1
x=xy^2\qquad\qquad\;\;\; (3)
Substitute the value of y into Eq. (3)
x=x(x-1)^2
x=x(x^2-2x+1)
x=x^3-2x^2+x
0=x^3-2x^2
0=x^2(x-2)
x=0 or x=2
We have two pairs (0,\textrm{-}1) and (2,1).
Suppose x-y<0
-x+y=1
y=x+1
Substitute the value of y into Eq. (3)
x=x(x+1)^2
x=x(x^2+2x+1)
x=x^3+2x^2+x
0=x^3+2x^2
0=x^2(x+2)
x=0 or x=\textrm{-}2
We have two more pairs (0,1) and (\textrm{-}2,\textrm{-}1).
Four different real-valued pairs (x,y) satisfy the system of two equations.

Answer: 4

Advertisements

About mvtrinh

Retired high school math teacher.
This entry was posted in Problem solving and tagged , , , . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s