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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.

Re: 152. Algebra Absolute value [#permalink]
10 Mar 2011, 10:19

3

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Expert's post

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.

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.

Re: For how many ordered pairs (x, y) that are solutions of the [#permalink]
05 Jul 2013, 02:54

2

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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)