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If x, y, and z are integers, y + z = 13, and xz = 9, which o
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30 Aug 2018, 07:50
30 Aug 2018, 07:50
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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
(A) x is even
(B) x = 3
(C) y is odd
(D) y > 3
(E) z < x
Re: If x, y, and z are integers, y + z = 13, and xz = 9, which o
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07 Sep 2018, 19:13
07 Sep 2018, 19:13
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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
(A) x is even
(B) x = 3
(C) y is odd
(D) y > 3
(E) z < x
xz = 9
Possible combinations:
3*3 = 9
1*9 = 9
9*1= 9
Now we can fint out the different values of y substituting the value of z.
y + z = 13
y + 3 = 13
y = 10
again,
y + z = 13
y + 1 = 13
y = 12
again,
y + z = 13
y + 9 = 13
y = 4.
In all cases y>3.
The best answer is D.
General Discussion
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Re: If x, y, and z are integers, y + z = 13, and xz = 9, which o
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09 Sep 2018, 09:47
09 Sep 2018, 09:47
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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
(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 INTEGERS
So, 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 = 4
CASE B: x = -1, z = -9, and y = 22
CASE C: x = 9, z = 1, and y = 12
CASE D: x = -9, z = -1, and y = 14
CASE E: x = 3, z = 3, and y = 10
CASE F: x = -3, z = -3, and y = 16
When 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
