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

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

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

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23 Oct 2016, 18:43
Thanks Karishma. Great post..

I was under the impression that when we say |X| < a, we get two options

x<a OR X >=-a..

How does the equal to sign work? Is the same sign preserved when the modulus sign is opened in both cases?

Another eq:

|X| <= a

X <=a OR X>-a... Will the it be >=-a ?? When does the equal to sign come..

In general I am confused about when we open the modulus, how does the equal to sign come in for both the positive and negative cases.. Any explanation will be really helpful!

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

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23 Oct 2016, 21:50
cuhmoon wrote:
Thanks Karishma. Great post..

I was under the impression that when we say |X| < a, we get two options

x<a OR X >=-a..

How does the equal to sign work? Is the same sign preserved when the modulus sign is opened in both cases?

Another eq:

|X| <= a

X <=a OR X>-a... Will the it be >=-a ?? When does the equal to sign come..

In general I am confused about when we open the modulus, how does the equal to sign come in for both the positive and negative cases.. Any explanation will be really helpful!

The equal to sign can be used in either case.

|x| = x if x >=0 and
|x| = -x if x < 0 (it can be x <= 0 too)

Note that when x = 0, then |x| = x and |x| = -x too since -0 = 0 only.
You can include it in either range. Both work.
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Re: For how many ordered pairs (x , y) that are solutions of the  [#permalink]

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24 Oct 2016, 13:42
Thanks Karishma. I do understand that but my doubt is whether the equal to sign comes for both the cases or any one?

If it just comes for once - then how come for |x| <=12, we get both cases -12 and +12

We should consider

X<=12
and
X>-12.. thus -12 should not be included if the equal to sign isn't considered ?? Please help me understand this..

Another example:

Will this open up as

x-4 <=9 and x-4>9

or
x-4<=9 and x-4>=-9

Will the greater/equal to come for both? or just for one case..

In this there is another question on

If x/|x| > x.. here you have solved the question but not using the = sign for any of the cases.. Why is that so?

In summary - i'm confused when does one use the = sign and when doesn't one?

VeritasPrepKarishma wrote:
cuhmoon wrote:
Thanks Karishma. Great post..

I was under the impression that when we say |X| < a, we get two options

x<a OR X >=-a..

How does the equal to sign work? Is the same sign preserved when the modulus sign is opened in both cases?

Another eq:

|X| <= a

X <=a OR X>-a... Will the it be >=-a ?? When does the equal to sign come..

In general I am confused about when we open the modulus, how does the equal to sign come in for both the positive and negative cases.. Any explanation will be really helpful!

The equal to sign can be used in either case.

|x| = x if x >=0 and
|x| = -x if x < 0 (it can be x <= 0 too)

Note that when x = 0, then |x| = x and |x| = -x too since -0 = 0 only.
You can include it in either range. Both work.
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Re: For how many ordered pairs (x , y) that are solutions of the  [#permalink]

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13 May 2017, 09:27
daviddaviddavid wrote:
i don't get it

is there a typo in the official guide.

In my book it stated |y|=12

It's $$|y| \leq 12$$, not |y| = 12. Please tell if there is anything else unclear?
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Re: For how many ordered pairs (x , y) that are solutions of the  [#permalink]

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13 May 2017, 09:33
Bunuel wrote:
daviddaviddavid wrote:
i don't get it

is there a typo in the official guide.

In my book it stated |y|=12

It's $$|y| \leq 12$$, not |y| = 12. Please tell if there is anything else unclear?

ok thanks

then there is a typo in the official guide math review. Ive got it right in front of me and it says |y| = 12

then id suggest x=(12-y)/2

thus, y is even

now select all even numbers from -12<y<12
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Re: For how many ordered pairs (x , y) that are solutions of the  [#permalink]

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15 May 2017, 11:01
o (Zero) is even or odd? Is Zero integer?
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Re: For how many ordered pairs (x , y) that are solutions of the  [#permalink]

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15 May 2017, 11:04
KARTAB wrote:
o (Zero) is even or odd? Is Zero integer?

ZERO:

1. 0 is an integer.

2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. 0 is neither positive nor negative integer (the only one of this kind).

4. 0 is divisible by EVERY integer except 0 itself.

Check more here: http://gmatclub.com/forum/number-proper ... 74996.html
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Re: For how many ordered pairs (x , y) that are solutions of the  [#permalink]

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31 May 2017, 04:41
Easy question for me, second equation gives the range of y as -12 to 12 both inclusive.

Insert even values of y in equation 1 and we'll get the integer values of x.

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

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02 Jun 2017, 10:54
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

This problem is best tackled via a table.

You can set the columns up as follows: x, 12 - 2x, |y|

x | 12 -2x| |y|
0 | 12 | 12
1 | 10| 10
2 ...
3
4
5
6
7
8
9
10
11
12

You'll note that the values here form a pattern. The values that satisfy |y| <= 12 are symmetric about 0, where x = 6. We have six values that satisfy the equation above point (6,0), so we must have six pairs of x-y coordinates below (6,0) that satisfy it as well.

So we'd expect 2*6 + 1 = 13 total values.
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Re: For how many ordered pairs (x , y) that are solutions of the  [#permalink]

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02 Aug 2017, 02:22
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

x and y are both integers.

$$|y| \leq 12$$
$$-12 \geq |y| \leq 12$$

Also 2x +y = 12
-> x = (12-y)/2 = 6-y/2

So, Y must be even
Hence Y can be -12,-10,-8,-6,-4,-2,0,2,4,6,8,10,12

Total 13 values

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

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30 Aug 2017, 06:26
|y| =< 12 => -12 =< y =<12
=> y= {-12, -11, -10,...,10,11,12}

2x + y=12 => x=(12-y)/2 => y is
even
Count y even from -12 to 12:
(12-(-12))/2 +1 = 13

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Re: For how many ordered pairs (x , y) that are solutions of the   [#permalink] 30 Aug 2017, 06:26

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