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

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How many ordered pairs (x , y) are solutions to the above equat [#permalink]
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BrentGMATPrepNow
The equation $$x + y = xy = \frac{x}{y}$$ has how many real solutions?

A) 0
B) 1
C) 2
D) 3
E) More than 3
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
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Re: How many ordered pairs (x , y) are solutions to the above equat [#permalink]
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
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Re: How many ordered pairs (x , y) are solutions to the above equat [#permalink]
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Re: How many ordered pairs (x , y) are solutions to the above equat [#permalink]
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