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

SOLUTION

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?

(A) 7
(B) 10
(C) 12
(D) 13
(E) 14

Given: \(-12\leq{y}\leq{12}\) and \(2x+y=12\)

Rearrange \(2x+y=12\) to get \(y=12-2x=2(6-x)=even\), (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.

P.S. The ordered pairs of (x, y)would be:
(12, -12)
(11, -10)
(10, -8)
(9, -6)
(8, -4)
(7, -2)
(6, 0)
(5, 2)
(4, 4)
(3, 6)
(2, 8)
(1, 10)
(0, 12)
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2x + y = 12 --> x= (12 - y)/2 = 6 - y/2. Thus every even value of y will yield integer value of x too.

|y| <= 12 --> There are 13 even values of y: 12 - (-12) = 24/2 + 1 = 13

Answer: D

|y| <= 12 means the range of Y is -12<=Y<=12
Let'ssimplify the first equation X=(12-y)/2 -> So in order both x and y to be an integer 12-y must be even.
We have 13 even numbers in the range of -12<=Y<=12: These are -12,-10,-8,-6,-4,-2,0,2,4,6,8,10,12 (don't forget to count 0 and 12)

Answer is (D).

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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The solution of \(|y| \leq 12\) is straight forward.
\(-12 \leq y \leq 12\)
(If you are not comfortable with this, check out my video: https://youtu.be/oqVfKQBcnrs )


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.

hello, my quant session continues

guys what does "many ordered pairs" mean ? i didnt understand the question itself. i thought it was coordinate geometry question

why are we looking into ODD and EVEN integers ?

can someone explain this please ?

Hi Dave,

Ordered pair means for what values of x and y the given condition satisfy.
Here we are discussing about the odd and even because from the first equation after simplifying further we can get x= 6-y/2.
So we have figure out for what values of y x is an integer.And from equation 2 we can get the values of y as -12<= y<=12.
So for x to be integer y has to an even integer( as only even integers are divisible by 2).so here our answer is to find how even integers are present between -12 and 12 i.e 13 .(don't forget to include 0).hope it helps

Sent from my XT1663 using GMAT Club Forum mobile app

For the inequality |y| ≤ 12, we see that -12 ≤ y ≤ 12

For the equation 2x + y = 12, we see that x = (12 - y)/2. If x has to be an integer, then y has to be an even integer; thus, y can be any of the even integers from -12 to 12, inclusive. Since there are

(12 - (-12))/2 + 1 = 24/2 + 1 = 13

even integers for y, there will be 13 corresponding integers for x. Hence, there are 13 ordered pairs (x, y) that are solutions to the system and where x and y are both integers.

Answer: D
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Given:
2x + y = 12
|y| <= 12

Asked: For how many ordered pairs (x , y) that are solutions of the system above are x and y both integers?

2x + y = 12 => y is even since both 2x and 12 are even
|y| = {0,2,4,6,8,10,12}

y={0,2,-2,4,-4,6,-6,8,-8,10,-10,12,-12}
x={6,5,7,4,8,3,9,2,10,1,11,0,12}
There are 12 (x,y) ordered pairs.

The ordered pairs of (x, y) are:

(12, -12)
(11, -10)
(10, -8)
(9, -6)
(8, -4)
(7, -2)
(6, 0)
(5, 2)
(4, 4)
(3, 6)
(2, 8)
(1, 10)
(0, 12)

IMO D

y=12-2x
y=2(6-x)

also |y| <= 12

so |2(6-x)| <= 12

i.e. -12 <= 2 (6-x) <= 12
i.e. -6 <= 6-x <= 6
i.e. -12 <= -x <= 0
i.e. 0<= x<= 12

So x can be any interger among integers from 0 to 12 .

Answer 13

VeritasKarishma JeffTargetTestPrep
I used this same process above ^^. I substituted x into |y| <= 12 to find that there are 13 values of x because the range was [0,12]
Is this a correct way to solve this problem? I see that it is different from your solutions...

Yes, this method is correct too. It uses algebraic substitution to solve.

Here :
2x+y=12
|y|≤12

implies -12 ≤ y ≤ 12 y can take, since it's given x n y are integers then , 2x = 12 -y so 12 -y needs to be multiple of 2.
If we subtract odd values from 12 it will give odd value, so need to avoid odd values of y between -12 and 12 which give us 12 even values.
Need to consider 0 as well.
So total values 13.
Ans D

1) 2 x + y = 12, from this equation we can find y = 12 - 2x
2) |y|≤12, from this inequality we can deduce following: -12 ≤ y ≤12
Take equation from 1st step and put into inequality in 2nd step and get following inequality: -12 ≤ 12-2x ≤12. Solve it as follow:
-12 ≤ 12-2x ≤12 |subtract 12 each side
-24≤ - 2x ≤ 0 |divide by -2 each side
12 ≥ x ≥ 0, so x can be anything from 0 to 12 including, and that is 13 numbers which is our answer choice D.

2x+y=12
|y|≤12

x and y are integers
2x+y=12
Here 2x and 12 are even
So y must be even
For every even y, you will definitely get a value of x
For example: y = -6, x = 9 | y = 6, x = 3, etc.
Total possible values of y are:
-12, -11, ... 0, ... 11, 12
Of these, the even ones are:

-12, -10, ... 0, ... 6, 8, 10, 12

Thus, there are 13 solution sets

Answer D

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