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For how many ordered pairs (x , y) that are solutions of the

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2x + y = 12
|y| <= 12

For how many ordered pairs (x , y) that are solutions of the system above are x and y both integers?

A. 7
B. 10
C. 12
D. 13
E. 14

OPEN DISCUSSION OF THIS QUESTION IS HERE: for-how-many-ordered-pairs-x-y-that-are-solutions-of-the-110687.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 09 Mar 2014, 23:57, edited 2 times in total.
Renamed the topic, edited the question and added the OA.

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New post 24 May 2007, 09:28
Can you explain why the Y values go from -12 to 12? I a little new at this forum. I originally got 12 total possibilities.

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2x + y = 12


|y| <= 12 means that y can be anything between 12 and -12 inclusive. Absolute values always indicates a range of numbers... this is the easy way to think about abs. values.

Ok, now you've narrowed down the answer choices to 25 possible numbers... which doesn't help you with the answers given. Next you need to find a way of eliminating more answer choices...

Simplify 2x + y = 12 to

x + y/2 = 3

Now looking at that, you know y has to be an even number to yield an integer... so the initial pool of 25 numbers is now narrowed down to 13, hence the answer.

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New post 24 May 2007, 19:57
This is a good question, coz I did not pay attention to the "(x,y) that will yield x and y to be integers" part of the question, so I was stuck with the answer being 25 and was stumped by the choices.
Good job you guys....i guess I should read the question clearly

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\(2x + y = 12\)
\(|y| \leq 12\)

For how many ordered pairs (x, y) that are solutions of the system above are x and y both integers?

A. 7
B. 10
C. 12
D. 13
E. 14

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Re: Quant Rev. #152 [#permalink]

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tonebeeze wrote:
152.

\(2x + y = 12\)
\(|y| \leq 12\)

For how many ordered pairs (x, y) that are solutions of the system above are x and y both integers?

a. 7
b. 10
c. 12
d. 13
e. 14


The solution of \(|y| \leq 12\) is straight forward.
\(-12 \leq y \leq 12\)
(If you are not comfortable with this, check out my blog post:
http://www.veritasprep.com/blog/2011/01/quarter-wit-quarter-wisdom-do-what-dumbledore-did/

If both x and y have to be integers, y should be an integer and hence can take any value from the set {-12, -11, -10 ... 10, 11, 12} i.e. any one of 25 values (these are 25 values -12 to -1 (12 values), 0, 1 to 12 (another 12 values)) 13 of them are even and 12 of them are odd.

\(2x + y = 12\)
Every time y is even, x will be integer. e.g. y = 12, x = 0 (because x = (12 - even)/2 will be an integer)
Every time y is odd, x will be non-integer e.g. y = 1, x = 5.5 (because x = (12 - odd)/2 will not be an integer)

Therefore, for 13 values, x and y both will be integers.
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Re: 152. Algebra Absolute value [#permalink]

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Sol:
|y| <= 12
Means;
-12<=y<=12

2x + y = 12
x = (12-y)/2

x will be integers for y=even; because even-even = even and even is always divisible by 2.

We need to find out how many even integers are there between -12 and 12

((12-(-12))/2)+1 = (24/2)+1 = 12+1 = 13

Ans: "D"
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Baten80 wrote:
2x + y = 12
|y| <= 12

152. For how many ordered pairs (x , y) that are solutions of the system above are x and y both integers?
A. 7
B. 10
C. 12
D. 13
E. 14


Given: \(-12\leq{y}\leq{12}\) and \(2x+y=12\) --> \(y=12-2x=2(6-x)=even\), (as \(x\) must be an integer). Now, there are 13 even numbers in the range from -12 to 12, inclusive each of which will give an integer value of \(x\).

Answer: D.
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-12<= y <=12
gives 0<=x <=12

thus 13 values in total.
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Re: PS-ordered pairs (x , y) [#permalink]

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Balvinder wrote:
2x + y = 12
|y| <= 12


For how many ordered pairs (x , y) that are solutions of the system above are x and y both integers?
A. 7
B. 10
C. 12
D. 13
E. 14


|y| <= 12
-12<=y<=12

2x + y = 12
x=(12-y)/2

To find number of integer pairs, we just need to find even number of y's, because even y will make "(12-y)" even as well and only even numbers divide by 2 evenly to give an integer.

e.g.
x=(12-y)/2; for y=1; x=(12-1)/2=11/2=5.5(Not an integer because y is odd)
x=(12-y)/2; for y=0; x=(12-0)/2=12/2=6(An integer because y is even)

Thus, if we find the number of even y's, we should be good.

-12<=y<=12
What is the first even number greater than or equal to -12?
-12
What is the last even number smaller than or equal to +12?
+12

\(Count=\frac{Last Even-First Even}{2}+1\)

\(Count=\frac{12-(-12)}{2}+1=12+1=13\)

Ans: "D"
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Re: PS-ordered pairs (x , y) [#permalink]

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Number of ordered pairs = number of integers between 12 and 0 (both inclusive)
= 13
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Re: Quant Rev. #152 [#permalink]

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New post 07 Sep 2012, 00:22
Hi Karishma

Using the number properties this indeed is very convenient to solve. I was wondering can we substitute y = 12 - 2x in the inequality and solve for the possible values of x.

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ratinarace wrote:
Hi Karishma

Using the number properties this indeed is very convenient to solve. I was wondering can we substitute y = 12 - 2x in the inequality and solve for the possible values of x.


Certainly and it is quick too.

y = 12 - 2x
Whenever x is an integer, y will be an integer. So if we can solve for integral values of x, the number of values we get will be the number of solutions.

\(|y| \leq 12\)

\(|12 - 2x| \leq 12\)

\(|x - 6| \leq 6\)

From 6, x should be at a distance less than or equal to 6. So x will lie from 0 to 12 i.e. 13 values. (Check http://www.veritasprep.com/blog/2011/01 ... edore-did/ if this is not clear)

There are 13 solutions.
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Re: For how many ordered pairs (x, y) that are solutions of the [#permalink]

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New post 25 Sep 2012, 02:50
Responding to a pm:
Changing the sign within the mod has no impact on anything outside the mod.

\(|6 - x| \leq 12\) is same as
\(|x - 6| \leq 12\)

Think about it: Whether you write |x| or |-x|, it is the same.
|6| = |-6|

So for every value of x,
|x - 6| = |6 - x|
So you don't need to flip the inequality sign.

|x - 6| and - |x - 6| are of course different. If you change |x - 6| to - |x - 6|, you will need to flip the inequality sign.
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\(y=12-2x=2*(6-x).\)
Since \(|y| \leq 12 , -12 \leq y \leq 12\) . Substituting for y from above, \(-6 \leq (6-x) \leq 6.\). This reduces to \(x \geq 0\) and \(x \leq 12.\) Including 0 and 12 there are thus 13 integer solutions.
Answer is (d)

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2x+y=12
|y|<=12

For how many ordered pairs (x, y) that are solutions of the system above are x and y both integers?

y=12-2x
|y|<=12
|12-2x| <= 12
12 - 2x <= 12
-2x <= 0
x>=0

-(12-2x) <= 12
-12+2x <= 12
2x <= 24
x<=12

13 solutions between 0 and 12 inclusive.

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2x + y = 12
|y| <= 12

For how many ordered pairs (x , y) that are solutions of the system above are x and y both integers?


A. 7
B. 10
C. 12
D. 13
E. 14

Given: \(-12\leq{y}\leq{12}\) and \(2x+y=12\) --> \(y=12-2x=2(6-x)=even\), (as \(x\) must be an integer). Now, there are 13 even numbers in the range from -12 to 12, inclusive each of which will give an integer value of \(x\).

Answer: D.

OPEN DISCUSSION OF THIS QUESTION IS HERE: for-how-many-ordered-pairs-x-y-that-are-solutions-of-the-110687.html
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Re: For how many ordered pairs (x, y) that are solutions of the [#permalink]

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New post 09 Apr 2016, 14:15
abs 12 = -12 all the way to 12, which is 25 integers including "0".

y= 2(6-x) = even.

So y = even. and y is equal the 25 number range. How many possible even numbers are in that range?

Answer is 13 possible even y numbers including zero "0".

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Re: For how many ordered pairs (x, y) that are solutions of the [#permalink]

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New post 12 Apr 2016, 00:23
tonebeeze wrote:
\(2x + y = 12\)
\(|y| \leq 12\)

For how many ordered pairs (x, y) that are solutions of the system above are x and y both integers?

A. 7
B. 10
C. 12
D. 13
E. 14


|y| <= 12 means range of y is -12 <= Y <= +12. which means Y can take any of the value in the set (-12, -11, -10......-1,0,1.....10,11,12).

now that we are given 2x + y = 12, y = 12 - 2x

we can include all the integer values for X as a solution for y = 12 - 2x as long as y falls in the above range mentioned. Such values of X are (0,1,2....12). 13 is the count for this set. Answer is D.
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