## 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: $4$