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Re: The equation x + y = xy = x/y has how many [#permalink]
2
vndnjn wrote:
why x = 0 and y = 0, not a solution??


Great question!

If x = 0 and y = 0, then x + y = 0, and xy = 0, but x/y does not equal 0.
The fraction 0/0 is undefined (just like the fraction 5/0 is undefined)
That is, 0/0 does not have any real numerical value.
As such. we can't say that 0/0 = 0

Cheers,
Brent
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Re: The equation x + y = xy = x/y has how many [#permalink]
2
x+y=xy=(x/y)

we break this into three separate equalities

(1) x+y=xy
(2) xy=(x/y)
(3) x+y=x/y

Remember we need solutions for that work for all (1),(2), and (3)

From (2) we get
xy^2=x
xy^2-x=0
x(y^2-1)=0
x=0, y=1, y=-1

From (1) we get

if x=0
y=0, but y can't be zero or else (x/y) would be undefined

if y=1
x+1=x
1=0, this is a contradiction, therefore y is not 1

if y=-1
x-1=-x
2x=1
x=1/2

So we have the solution x=1/2, y=-1

Let's see if it finally works in (3)


(1/2)*-1=(1/2)/(-1)
-1/2=-1/2
This is true.


Therefore we have that x=1/2 and y=-1 is the only solution that works

Final Answer: B
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Re: The equation x + y = xy = x/y has how many [#permalink]
Thank you for explaining! My only question is, why is the answer only ONE solution. Wouldn't it be two solutions, x = 0.5, and y = -1? THANK YOU! Or are negatives not considered "real" solutions?


GreenlightTestPrep wrote:
GreenlightTestPrep wrote:
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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The equation x + y = xy = x/y has how many [#permalink]
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0 makes the fraction UNDEFINED so it is not possible

1 makes the equation x+1=x

x-x+1=0

0+1=0

1=0 so it is not possible

-1 is our value

if you insert -1 and obtain a cogent value (0.5) so our equation has ONE solution valid
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Re: The equation x + y = xy = x/y has how many [#permalink]
Thanks to you all for the solution.

What concept(s) does this question test?

I can see divisibility/multiples/fractions but it seems to be testing systems of equations, silmultaneous equations etc.

Please help.
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The equation x + y = xy = x/y has how many [#permalink]
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Re: The equation x + y = xy = x/y has how many [#permalink]
Thanks so much!

I will look at them.
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Re: The equation x + y = xy = x/y has how many [#permalink]
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