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Bunuel
Could you change the option (A) to reflect exactly what Quant Review text has ?
The above screenshot taken from Quant Review 2025-26.
parkhydel
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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GMAT-Club-Forum-qgminx21.png
GMAT-Club-Forum-qgminx21.png [ 17.33 KiB | Viewed 520 times ]
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kingbucky

Bunuel
Could you change the option (A) to reflect exactly what Quant Review text has ?
The above screenshot taken from Quant Review 2025-26.
parkhydel
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
Attachment:
GMAT-Club-Forum-qgminx21.png

Edited the option. Thank you! +1
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I didn't do much calculation in this, considering i went through with logic here, if xy is greater than 0, both can be -ve or +ve and not one can be 0, both can't be negative because xy=x+y will give us a negative value if both are negative, so in order to satisfy both, both have to be positive. now if we consider 0.2 and 0.3, the point moves 2 places when we multiply, whereas one if we add, so x has to be greater than 1?
parkhydel
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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Here is a more simpler way to do it.
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Strange, this is classified as easy...
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Essentially, there is no special formula/concept.
It is mainly an algebra + sign reasoning question:
xy > 0
=> x, y have the same sign
Also:
xy = x + y > 0
If x and y were both negative:
x + y < 0
But xy = x + y > 0
Contradiction.
Therefore:
x > 0 and y > 0
Now manipulate:


xy = x + y

Divide by y:

x = x/y + 1

Since:

x > 0 and y > 0

=> x/y > 0

Therefore:

x = 1 + positive number

=> x > 1

-----

Way 2: Division Method
xy = x + y
Divide by y:
x = x/y + 1
Since x > 0 and y > 0:
x/y > 0
Therefore:
x > 1
-----

Way 3: Plug Values
Try x = 1/2:
(1/2)y = 1/2 + y
y = 1 + 2y
y = -1
xy < 0
Does not work.
Try x = 2:
2y = 2 + y
y = 2
xy = 4 > 0
Works.
Therefore:
x > 1
-----

Best method:
xy = x + y
x = x/y + 1
x/y > 0
Therefore x > 1.
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