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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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New post 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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New post 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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New post 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:

How about |x-4|<9

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..

Another example is here: inequalities-made-easy-206653.html#p1582891

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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New post 25 Oct 2016, 03:44
1
cuhmoon wrote:
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:

How about |x-4|<9

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..

Another example is here: inequalities-made-easy-206653.html#p1582891

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.


Include the "=" sign in either range - it won't make a difference. The point is that you need to consider all possible values of x. So you need to consider what is the value of |x| when x is positive, when x is 0 and when x is negative. So you take the cases as x >= 0, x < 0 OR as x > 0, x <= 0 - it won't make any difference. No point including 0 in both ranges. Once you consider it in one range, it is sufficient.

Imagine a case where you are given that x is non-positive. In that case, you need to say
|x| = -x when x <= 0
You need to consider x = 0 as well and you can do that along with x < 0 itself. Because x can take the value 0 too, you cannot ignore it and use only x < 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 13 May 2017, 09:27
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Re: For how many ordered pairs (x , y) that are solutions of the  [#permalink]

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

can someone help please ?

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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New post 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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New post 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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New post 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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New post 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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New post 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

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

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New post 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

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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New post 02 Sep 2017, 07:08
1
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


From the inequality |y| ≤ 12, we see that -12 ≤ y ≤ 12. If we want y to be an integer, then y is any integer from -12 to 12 inclusive. Since we want x to be an integer also, let’s isolate x in terms of y:

2x + y = 12

2x = 12 - y

x = 6 - y/2

We see that in order for x to be an integer, y must be even so that y/2 is an integer. Thus, y is any of the integers in the following list:

-12, -10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10, 12

For any of the 13 integers in the above list, x will be also an integer. Thus, there are 13 ordered pairs that satisfy the system with both x and y being integers.

Answer: D
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Re: For how many ordered pairs (x , y) that are solutions of the   [#permalink] 02 Sep 2017, 07:08

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