How many ordered pairs (x , y) are solutions to the above equat

\(xy = \frac{x}{y}\)

\(xy - \frac{x}{y} = 0\)

\(x(y-\frac{1}{y}) = 0\)

\(\frac{x(y^2-1)}{y} = 0\)

Either x=0 or y = 1 or -1

Case 1- x=0

x+y=xy
y= 0

But \(\frac{x}{y}\) is undefined at y= 0

Reject this case

Case 2- y= 1

x+y=xy
x+1= x
1=0
Reject this one too

Case 3- y= -1

x+y=xy
x-1= -x

\(x=\frac{1}{2}\)

At \(x=\frac{1}{2}\) and \(y=-1\)

\(x+y = -\frac{1}{2}\); \(xy=\frac{-1}{2}\); \(\frac{x}{y}= \frac{-1}{2}\) [We got one]

B

To begin, if \(xy = \frac{x}{y}\), we can conclude that \(y\neq0\), otherwise \(\frac{x}{y}\) is undefined.

Take: \(xy = \frac{x}{y}\)
Multiply both sides of the equation by \(y\) to get: \(xy^2 = x\)
Subtract \(x\) from both sides to get: \(xy^2 - x = 0\)
Factor to get: \(x(y^2 - 1) = 0\)
Factor again to get: \(x(y - 1)(y + 1) = 0\)
There are 3 POSSIBLE solutions to the above equation: \(x = 0\), \(y = 1\) and \(y = -1\)

Let's examine each possible case:

case i: \(x = 0\)
Substitute this into the original equation to get: \(0 + y = (0)y = \frac{0}{y}\)
For this equation to hold true, we need \(y = 0\), but we already showed that \(y\) cannot equal \(0\)
So, it can't be the case that \(x = 0\)

case ii: \(y = 1\)
Substitute this into the original equation to get: \(x + 1 = x(1) = \frac{x}{1}\)
Simplify: \(x + 1 = x = x\)
Since there are no solutions to the equation \(x + 1 = x\), it can't be the case that \(y = 1\)

case iii: \(y = -1\)
Substitute this into the original equation to get: \(x + (-1) = x(-1) = \frac{x}{-1}\)
Simplify: \(x - 1 = -x = -x\)
Now take: \(x - 1 = -x\)
Add \(x\) to both sides: \(2x - 1 = 0\)
Add \(1\) to both sides: \(2x = 1\)
Solve: \(x = 0.5\)
So, \(x = 0.5\) and \(y = -1\) is the ONLY possible solution to the given equation.

Answer: B

Cheers,
Brent

Asked: The equation \(x + y = xy = \frac{x}{y}\) has how many real solutions?

x + y = xy
x + y - xy = 0
x(1-y) = - y
x = y/(y-1)

xy = x/y
x(y-1/y) = 0
y - 1/y = 0
y^2 = 1
y = 1 or -1

If y=1;x is undefined.
If y=-1; x = 1/2

IMO B