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What values of x have a corresponding value of y that satisfies both x

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What values of x have a corresponding value of y that satisfies both x  [#permalink]

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New post 27 Apr 2020, 13:29
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What values of x have a corresponding value of y that satisfies both xy > 0 and xy = x + y ?


A. \(x \leq 1\)

B. \(-1 < x \leq 0\)

C. \(0 < x \leq 1\) 

D. \(x > 1\)

E. All real numbers


PS22680.02
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What values of x have a corresponding value of y that satisfies both x  [#permalink]

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New post 27 Apr 2020, 13:51
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xy = x + y

xy-y = x

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

xy> 0

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

\(x^2\) is always greater than 0; hence x-1>0 or x>1.

D




parkhydel wrote:
What values of x have a corresponding value of y that satisfies both xy > 0 and xy = x + y ?


A. \(x \leq 1\)

B. \(-1 < x \leq 0\)

C. \(0 < x \leq 1\) 

D. \(x > 1\)

E. All real numbers


PS22680.02
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What values of x have a corresponding value of y that satisfies both x  [#permalink]

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New post Updated on: 28 May 2020, 10:12
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parkhydel wrote:
What values of x have a corresponding value of y that satisfies both xy > 0 and xy = x + y ?

A. \(x \leq 1\)
B. \(-1 < x \leq 0\)
C. \(0 < x \leq 1\) 
D. \(x > 1\)
E. All real numbers

PS22680.02


Given: \(xy = x + y\)
Subtract \(y\) from both sides to get: \(xy - y= x\)
Factor: \(y(x-1)=x\)
Divide both sides by \(x-1\) to get: \(y=\frac{x}{x-1}\)

Important: At this point, there are at least two different approaches we can take

APPROACH #1:
Now that we know \(y=\frac{x}{x-1}\), we can readily see that x cannot equal 1, otherwise, the denominator in y is 0, which would make y undefined.
At this point we can eliminate answer choices A, C and E, since they all allow for x to equal 1.
Since we're down to just two answer choices, let's just test some x-value.

For example, if \(x = 2\), then we get: \(y=\frac{x}{x-1}=\frac{2}{2-1}=2\).
So, \(x = 2\) and \(y = 2\) is a possible solution to the system of equations.
Since it's possible for x to equal 2, we can eliminate answer choice B

By the process of elimination, the correct answer is D.

APPROACH #2:
We've already concluded that: \(y=\frac{x}{x-1}\)

Now take the given inequality, \(xy > 0\), and replace \(y\) with \(\frac{x}{x-1}\) to get: \((x)(\frac{x}{x-1})>0\)
Simplify: \(\frac{x^2}{x-1}>0\)

Since \(x^2\) is always greater or equal to 0, it must be the case that the denominator, \(x-1\), is positive
In other words, it must be the case that: \(x-1>0\)
Add \(1\) to both sides of the inequality to get: \(x>1\)

Answer: D

Cheers,
Brent
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Originally posted by BrentGMATPrepNow on 27 Apr 2020, 20:48.
Last edited by BrentGMATPrepNow on 28 May 2020, 10:12, edited 1 time in total.
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Re: What values of x have a corresponding value of y that satisfies both x  [#permalink]

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New post 27 Apr 2020, 20:52
parkhydel wrote:
What values of x have a corresponding value of y that satisfies both xy > 0 and xy = x + y ?


A. \(x \leq 1\)

B. \(-1 < x \leq 0\)

C. \(0 < x \leq 1\) 

D. \(x > 1\)

E. All real numbers


PS22680.02


Asked: What values of x have a corresponding value of y that satisfies both xy > 0 and xy = x + y ?

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

Since xy > 0
x^2/(x-1)>0
\(x \neq 0\)
x-1>0
x>1

IMO D
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Re: What values of x have a corresponding value of y that satisfies both x  [#permalink]

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New post 26 May 2020, 11:26
nick1816 wrote:
xy = x + y

xy-y = x

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

xy> 0

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

x^2 is always greater than 0; hence x-1>0 or x>1

D

U make it look so easy. Your reasoning is amazing.Thanks



parkhydel wrote:
What values of x have a corresponding value of y that satisfies both xy > 0 and xy = x + y ?


A. \(x \leq 1\)

B. \(-1 < x \leq 0\)

C. \(0 < x \leq 1\) 

D. \(x > 1\)

E. All real numbers


PS22680.02
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What values of x have a corresponding value of y that satisfies both x  [#permalink]

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New post 28 May 2020, 09:55
1
BrentGMATPrepNow wrote:
parkhydel wrote:
What values of x have a corresponding value of y that satisfies both xy > 0 and xy = x + y ?

A. \(x \leq 1\)
B. \(-1 < x \leq 0\)
C. \(0 < x \leq 1\) 
D. \(x > 1\)
E. All real numbers

PS22680.02


Given: \(xy = x + y\)
Subtract \(y\) from both sides to get: \(xy - y= x\)
Factor: \(y(x-1)=x\)
Divide both sides by \(x-1\) to get: \(y=\frac{x}{x-1}\)

Important: At this point, there are at least two different approaches we can take

APPROACH #1:
Now that we know \(y=\frac{x}{x-1}\), we can readily see that x cannot equal 1, otherwise, the denominator in y is 0, which would make y undefined.
At this point we can eliminate answer choices A, C and E, since they all allow for x to equal 1.
Since we're down to just two answer choices, let's just test some x-value.

For example, if \(x = 2\), then we get: \(y=\frac{x}{x-1}=\frac{2}{2-1}=1\).
So, \(x = 2\) and \(y = 1\) is a possible solution to the system of equations.
Since it's possible for x to equal 2, we can eliminate answer choice B

By the process of elimination, the correct answer is D.

APPROACH #2:
We've already concluded that: \(y=\frac{x}{x-1}\)

Now take the given inequality, \(xy > 0\), and replace \(y\) with \(\frac{x}{x-1}\) to get: \((x)(\frac{x}{x-1})>0\)
Simplify: \(\frac{x^2}{x-1}>0\)

Since \(x^2\) is always greater or equal to 0, it must be the case that the denominator, \(x-1\), is positive
In other words, it must be the case that: \(x-1>0\)
Add \(1\) to both sides of the inequality to get: \(x>1\)

Answer: D

Cheers,
Brent


Hi BrentGMATPrepNow,

Thank you for the solution.
Just wanted to point out a small calculation error in the highlighted part. y should be 2.
Could you please amend the solution.

Regards,
Udit
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Re: What values of x have a corresponding value of y that satisfies both x  [#permalink]

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New post 28 May 2020, 10:13
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Top Contributor
uc26 wrote:

Hi BrentGMATPrepNow,

Thank you for the solution.
Just wanted to point out a small calculation error in the highlighted part. y should be 2.
Could you please amend the solution.

Regards,
Udit


Thanks for the heads up!!
I have edited my solution accordingly.

Kudos for you!!!!

Cheers,
Brent
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Re: What values of x have a corresponding value of y that satisfies both x  [#permalink]

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New post 28 May 2020, 11:10
xy = x+y
xy - y = x
or y = x/(x-1)

Given:: xy>0
==> x*x/(x-1) > 0
==> x^2 / (x-1) > 0

x^2 will always be +ve

so eqn will be +ve when x-1>0
or x>1

=> D
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Re: What values of x have a corresponding value of y that satisfies both x   [#permalink] 28 May 2020, 11:10

What values of x have a corresponding value of y that satisfies both x

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