sandy wrote:

If x, y, and z are integers, y + z = 13, and xz = 9, which of the following must be true?

(A) x is even

(B) x = 3

(C) y is odd

(D) y > 3

(E) z < x

The most limiting fact is that x, y, and z are

INTEGERSSo, if xz = 9, there are only 6 possible cases:

CASE A: x = 1 and z = 9

CASE B: x = -1 and z = -9

CASE C: x = 9 and z = 1

CASE D: x = -9 and z = -1

CASE E: x = 3 and z = 3

CASE F: x = -3 and z = -3

Since we also know that y + z = 13, we can add the corresponding

y-value to each of the 6 possible CASES to get:

CASE A: x = 1, z = 9, and y =

4CASE B: x = -1, z = -9, and y =

22CASE C: x = 9, z = 1, and y =

12CASE D: x = -9, z = -1, and y =

14CASE E: x = 3, z = 3, and y =

10CASE F: x = -3, z = -3, and y =

16When we scan the answer choices, we can see that D must be true since all 6 possible

y-values are greater than 3.

Answer: D

Cheers,

Brent

_________________

Brent Hanneson – Creator of greenlighttestprep.com

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