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

Question

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

Bunuel · Accepted Answer

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)

NoHalfMeasures · Answer

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

BrainLab · Answer

|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).

JeffTargetTestPrep · Answer

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

KarishmaB · Answer

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.

dave13 · Answer

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 ?

MaheshDuggirala · Answer

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

JeffTargetTestPrep · Answer

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

Kinshook · Answer

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

SHUBHAM GAUTAM · Answer

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

dc2880 · Answer

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   
Is this a correct way to solve this problem? I see that it is different from your solutions...

KarishmaB · Answer

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

sahuanamika · Answer

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

Vesper2018 · Answer

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.

sujoykrdatta · Answer

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

Posted from my mobile device

1616 · Answer

Hello  Bunuel ,

What is the difference between ordered and unordered pairs and how this question will have been different if it has asked about unordered pairs?

Bunuel · Answer

The difference between ordered and unordered pairs is in the sequence and significance of the elements:

Ordered Pairs : The sequence of elements matters. For example, (x=1, y=2) is different from (x=2, y=1) unless x equals y. Each specific pair is counted as unique based on the order of elements. 
  Unordered Pairs : The sequence of elements does not matter. For example, (x=1, y=2) is considered the same as (x=2, y=1).

For this particular question, the number of unordered pairs would not be different from the number of ordered pairs. But if there were a solution such as (4, 6) and (6, 4) satisfying the conditions, those would be counted separately as ordered pairs because (x=4, y=6) is different from (x=6, y=4).

However, if the question were to find the number of ordered pairs of (x, y) where both x and y are positive integers such that x + y = 4, the answer would be:

Ordered pairs : (1, 3), (2, 2), and (3, 1). This gives us 3 ordered pairs. 
  Unordered pairs : (1, 3) and (2, 2). Since (1, 3) and (3, 1) are considered the same pair, the correct unordered pairs are (1, 3) and (2, 2). This gives us 2 unordered pairs.

rguerrerogzz · Answer

I solved it this way. Please give feedback of my thinking process! Is it valid?

purvagupta · Answer

Bunuel , why is 0 considered an integer value when counting the amount of solutions here?

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards