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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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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 \(x\).

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.

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.

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05 Jul 2013, 02:54

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

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11 Apr 2016, 23:23

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, 17:43

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?

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.

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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