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If x and y are integers such that (5x + 7)y < 0 and (5x + 9)(y - 2) > 23 Feb 2026, 08:21
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E
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Difficulty:

95% (hard)

Question Stats:

38% (02:29) correct 62% (02:46) wrong based on 68 sessions

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If x and y are integers such that (5x + 7)y < 0 and (5x + 9)(y - 2) > 0, which of the following must be true?

I. y = 1
II. x < 0
III. xy < 0

A) I only
B) II only
C) III only
D) I and II only
E) I, II, and III
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E
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If x and y are integers such that (5x + 7)y < 0 and (5x + 9)(y - 2) > 26 Feb 2026, 01:28
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If x and y are integers such that (5x + 7)y < 0 and (5x + 9)(y - 2) > 26 Feb 2026, 02:18
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How about a clue? What must be true about x for 5X +7 and 5X +9 to have opposite signs?
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If x and y are integers such that (5x + 7)y < 0 and (5x + 9)(y - 2) > 09 Mar 2026, 08:18
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The key here is to realize that 5X +7 and 5X +9 must have the same sign if X is an integer . Thus y and y-2 have opposite signs. If I had dollar every time I I saw a student gloss over the word integer, right now I would be sipping something decadent somewhere exotic
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If x and y are integers such that (5x + 7)y < 0 and (5x + 9)(y - 2) > 10 Mar 2026, 21:17
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Therefore (5x +7)(5x + 9)(y)(y - 2)> 0
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If x and y are integers such that (5x + 7)y < 0 and (5x + 9)(y - 2) > 11 Mar 2026, 18:35
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Case 1

5x + 7 > 0

x > -7/5

or

y < 0



Case 2

5x + 7 < 0

x < - 7/5

or

y > 0

Case 3

5x + 9 > 0

x > - 9/5

or

y > 2

Case 4

5x + 9 < 0

x < - 9/5

or

y < 2



Case 1 and 3 does not provide value for y

Case 1 and 4 does not provide value for x

Case 2 and 3 does not provide value for x

Then lastly Case 2 and 4 tells us x is negative and y = 1

So answer is E

Adewale Fasipe, GMAT quant instructor, Lagos Nigeria.
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If x and y are integers such that (5x + 7)y < 0 and (5x + 9)(y - 2) > 11 Mar 2026, 23:04
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(5x+7)y < 0
Case 1a: y > 0; x < -7/5 = -1.4<0
Case 1b: y < 0; x > -7/5 = -1.4

(5x+9)(y-2) > 0
Case 2a: y > 2; x > -9/5 = -1.8
Case 2b: y < 2; x < -9/5 = -1.8<0

Case 1a + Case 2a:
y > 2; -1.8 < x < -1.4; Not feasible since there is no integer between -1.8 and -1.4

Case 1a + Case 2b:
0 < y < 2;
y = 1; Since y is integer between 0 & 2
x < -1.8 < 0
xy < 0

Case 1b + Case 2a:
y < 0 & y > 2; Not feasible

Case 1b + Case 2b:
y < 0; -1.4 < x < -1.8; Not feasible since there is no integer between -1.8 and -1.4 and -1.8 < -1.4

Only Case 1a + Case 2b is feasible with
y = 1
x< 0
and xy < 0

I. y=1
II. x< 0
III. xy < 0

All MUST BE TRUE

IMO E
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