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

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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
[Reveal] Spoiler: OA
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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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Re: 152. Algebra Absolute value [#permalink]

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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: 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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Re: Quant Rev. #152 [#permalink]

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

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

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New post 13 May 2017, 09:19
i don't get it :(

can someone help please ?

is there a typo in the official guide.

In my book it stated |y|=12
For how many ordered pairs (x, y) that are solutions of the   [#permalink] 13 May 2017, 09:19

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