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Manager  G
Joined: 03 Jun 2019
Posts: 79
What values of x have a corresponding value of y that satisfies both x  [#permalink]

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19 00:00

Difficulty:   55% (hard)

Question Stats: 63% (02:08) correct 37% (02:16) wrong based on 238 sessions

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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
DS Forum Moderator V
Joined: 19 Oct 2018
Posts: 1980
Location: India
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8
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
##### General Discussion
GMAT Club Legend  V
Joined: 11 Sep 2015
Posts: 4953
GMAT 1: 770 Q49 V46
What values of x have a corresponding value of y that satisfies both x  [#permalink]

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2
Top Contributor
1
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: $$x>1$$

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.
CEO  V
Joined: 03 Jun 2019
Posts: 3182
Location: India
GMAT 1: 690 Q50 V34 WE: Engineering (Transportation)
Re: What values of x have a corresponding value of y that satisfies both x  [#permalink]

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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

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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Kinshook Chaturvedi
Email: kinshook.chaturvedi@gmail.com
Intern  B
Joined: 18 Jun 2018
Posts: 27
Location: India
Schools: ISB
WE: Corporate Finance (Retail Banking)
Re: What values of x have a corresponding value of y that satisfies both x  [#permalink]

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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 S
Joined: 19 Sep 2018
Posts: 82
What values of x have a corresponding value of y that satisfies both x  [#permalink]

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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: $$x>1$$

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
GMAT Club Legend  V
Joined: 11 Sep 2015
Posts: 4953
GMAT 1: 770 Q49 V46
Re: What values of x have a corresponding value of y that satisfies both x  [#permalink]

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1
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

I have edited my solution accordingly.

Kudos for you!!!!

Cheers,
Brent
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IESE School Moderator S
Joined: 11 Feb 2019
Posts: 271
Re: What values of x have a corresponding value of y that satisfies both x  [#permalink]

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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
_________________

Cheers,
NJ 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  