If x and y are integers such that |y + 3| 3 and 2y 3x + 6 = 0,

Question

If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, what is the least possible value of the product xy?

(A) -12
(B) -3
(C) 0
(D) 2
(E) None of the above

This is   Question 5   of    the e-GMAT Question Series on Absolute Value.

Provide your solution below. Kudos for participation. Happy Solving! :-D

Best Regards
The e-GMAT Team

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EgmatQuantExpert · Accepted Answer

Official Explanation

Correct Answer: C

|y + 3| ≤ 3

means that the distance of y from the point -3 on the number line is less than, or equal to, 3

This gives us, -6 ≤ y ≤ 0

So, possible values of y (keeping in mind the constraint that y is an integer): -6, -5, -4, -3, -2, -1, 0

We’re given 2y – 3x = -6

So,  x= \frac{(2y+6)}{3}

For x to be an integer, y must be a multiple of 3.

This means, possible values of y:  -6 ,  -3 , 0
Corresponding values of x:  -2 ,  0 , 2
Corresponding values of xy:  12 ,  0 , 0

Thus, minimum possible value of xy: 0

UJs · Answer

Eq: 1   |y + 3| ≤ 3 and  Eq: 2   2y – 3x + 6 = 0, x & y are  integers 
Solve   Eq: 1    |y + 3| ≤ 3

+(y + 3) ≤ 3 
 y≤ 0

-(y + 3) ≤ 3 
 y\geq{-6}

Solve   Eq: 2    2y – 3x + 6 = 0

x = \frac{(2y + 6)}{3}

as range of y:  -6=<y<=0

we get 3 integer value of x , as x & y are integers for  y = {-6, -3, 0} x = {-2, 0, 0} & xy = {12, 0, 0}

lowest  xy = 0

Ans   : C

reto · Answer

How to deal with inequalities involving absolute values? First example shows us the so called "number case" In this case we have |y + 3| ≤ 3 which is generalized |something| ≤ some number. First we solve as if there were no absolute value brackets: y + 3 ≤ 3 y ≤ 0 So y is 0 or negative Second scenario - remove the absolute value brackets. Put a negative sign around the other side of the inequality, AND flip the sign: y + 3 >= -3 y >= -6 Therefore we have a possible range for y: -6=

EgmatQuantExpert · Answer

Dear reto Very well explained! Kudos for that.

Just want to ask 2 questions: 1. Please refer to the orange part in your solution. Is x = 2/3 a valid value for x? Why/ why not? 2. Is the question asking you to find the value of the product xy? Based on the answers to these 2 questions, you may feel a need to edit the last part of your solution :wink:

Best Regards Japinder

Sidhrt · Answer

Answer should be E.

As per the equation |y+3| <=3, y can have value from -6 to 0. 
Now as per the equation x=(2y+6)/3; x will have negative values for all y in {-6,-5,-4,-3} hence the xy will be (-x)*(-y) a positive value. for y in { -2,-1,0}, x will have { 2/3,4/3,2}. Hence x*y will have minimum at x=2/3,y=-2 or x=4/3,y=-1

csirishac · Answer

-6<=y<=0
To get the least value of xy, we have to take the least values of both the terms. Taking y=-6, x is -2. Xy is 12. Hence, none.

Note : It is clearly mentioned that both x and y are integers.

EgmatQuantExpert · Answer

Dear csirishac You bring up an interesting point by saying the part in red.

This part is not true. As you yourself saw, when you minimized the values of both x and y (both negative), you got a positive product. However, if x = +2 and y = 0 (both are actually the maximum values of x and y in this question), the product = 0, far lesser than the product 12 that you got above. So, be careful about what you need to minimize. Best Regards Japinder

kanigmat011 · Answer

If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, what is the least possible value of the product xy?

(A) -12
(B) -3
(C) 0
(D) 2
(E) None of the above

solving |y+3|<=3
gives 
 -6<=Y<=0

further equation 2y-3x+6=0
In order to fiind least value of x we nned to maximize y i,e 0
this gives x =2
And least value of y=-6

So product will be -12
 Bunuel  @VeritasPrepKarsihma  Engr2012  
Experts please let me know where I am missing something?

Bunuel · Answer

The relationship between the values of x and y are given by the formula 2y – 3x + 6 = 0. You cannot take the values of x and y separately, meaning that you cannot take x= 2 and y = -6, because if x = 2, then y = 0, and xy = 0.

elisabettaportioli · Answer

Japinder,
I'm getting back to yours qs:
1) x=2/3 must be rejected as per costraints in the question stem.
2) we are asked to find the least product of xy. This happens when y=0 x=2 or y=-3 x=0.

Thanks.

robu · Answer

is it that we should multiply corresponding pair of XY or can we multiply y= -6 and x= 2 XY =-12

pls. clear the doubt. thank in advance.

chetan2u · Answer

Hi,
the eq ly+3l<=3 gives following values of y..
y=0 ,-1,-2,-3,-4,-5,-6..
for these values x=2,-,-,0,-,-,-2...as x has to be an integer..
you have to take the corresponding values only..
i)when y=-6, then x =-2
ii)and when y=0, x=2..
iii) when y=-3, x=0..

ofcourse x as 0 and y as 0 is wrong inputs
so you get xy as 12 in (i) case and 0 in (ii) and (iii) case..
so 0 is the answer..
hope it helps

mvictor · Answer

we are given that
-3<= y <= -1.
y can take 3 values:
-3, -2, -1, and 0.

2y-3x +6 = 0
2y+6 = 3x
we can test values, but we can clearly see that if y is not -3/0, then x can't be an integer, but we need it to be an integer.
therefore, the only cases that actually work are y=-3, x=0 or y=0, and x=2
xy = 0 in any case

C is the answer.

Madhavi1990 · Answer

A simple algebraic approach.
We have |y + 3| less than eq to 3 which gives us -6 <=y<=0. So we have two values which are definitely for y (-6 and 0). Now we find x.
2y - 3x + 6 = 0 => 2y = 3x - 6 => y = 3x - 6/2

If we take y = -6; then we have :
- 6 = 3x - 6/ 2
-12 = 3x - 6
-6 = 3x
There fore x = -2.
X * Y = -2*-6 = positive 12 --> not the least value

Now lets take y = 0
0 = 3x - 6/2
3x = -6 => 6 = 3x so x =2
X*Y = 2*0 = 0

Would this approach be right? Expert comment would be welcome!

rocko911 · Answer

Yes Indeed....This approach is correct

luffy_ueki · Answer

If you visualize |y+3|<=3 on the number line you will get to know that y can take values in the range

Next express the give relation in terms of x i.e. y= (2/3)*(y-3).
and since y is in the range of   and not in (0,3) .. x will follow y's sign.
Thus, the product xy has to be >=0. 
Minumum being 0 .. check whether one can be 0 .. y=0 falls in the range. min. value of the product =0

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If x and y are integers such that |y + 3| 3 and 2y 3x + 6 = 0,

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