Baldini wrote: ↑

Thu May 14, 2009 2:34 am

If x and y are integers and 2x–y= 11, then 4x+ y cannot be:

(A) –5
(B) 1
(C) 13
(D) 17
(E) 55

OA is D . CAn someone please explain how to solve this quickly?
Thanks[/spoiler]

Solution:

Let 4x + y = z. If we add this with 2x - y = 11, we have: 6x = z + 11. Since x is an integer, we see that z + 11 has to be a multiple of 6.

If we multiply 2x - y = 11 by 2, we have 4x - 2y = 22. If we subtract this from 4x + y = z, we have 3y = z - 22. Since y is an integer, we see that z - 22 has to be a multiple of 3.

Now, let’s look at the given answer choices:

A. -5

If z = -5, we see that -5 + 11 = 6 is a multiple of 6 and -5 - 22 = -27 is a multiple of 3. So 4x + y can be -5.

B. 1

If z = 1, we see that 1 + 11 = 12 is a multiple of 6 and 1 - 22 = -21 is a multiple of 3. So 4x + y can be 1.

C. 13

If z = 13, we see that 13 + 11 = 24 is a multiple of 6 and 13 - 22 = -9 is a multiple of 3. So 4x + y can be 13.

D. 17

If z = 17, we see that 17 + 11 = 28 is NOT a multiple of 6. So 4x + y CAN’T be 17.

Alternate Solution:

Let’s rewrite the first equation as:

2x - y = 11

2x = 11 + y

4x = 22 + 2y

Now substitute 22 + 2y for 4x into the expression 4x + y, obtaining:

22 + 2y + y

22 + 3y

Let’s set 22 + 3y = q and see if we obtain an integer solution for y as we test each of the answer choices.


Choice A. (q = -5): 22 + 3y = -5 becomes 3y = 27. Thus, y = 9, which is an integer.

Choice B. (q = 1): 22 + 3y = 1 becomes 3y = -21. Thus, y = -7, which is an integer.

Choice C. (q + 13): 22 + 3y = 13 becomes 3y = 9. Thus, y = 3, which is an integer.

Choice D. (q = 17): 22 + 3y = 17 becomes 3y = -5. Thus, y = -5/3, which is NOT an integer.

Choice E. (q = 55): 22 + 3y = 55 becomes 3y = 33. Thus, y = 11, which is an integer.

Thus, the correct answer is D.

Answer: D