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

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If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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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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Originally posted by EgmatQuantExpert on 20 May 2015, 10:28.
Last edited by EgmatQuantExpert on 07 Aug 2018, 06:17, edited 3 times in total.
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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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New post 22 May 2015, 07:55
2
4
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
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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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New post Updated on: 21 May 2015, 14:20
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2
EgmatQuantExpert wrote:
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. The Official Answer and Explanation will be posted on 22nd May.

Till then, Happy Solving! :-D

Best Regards
The e-GMAT Team


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=<y<=0

Ok, so far so good, we're half way through. What about x?

Here's the formula: 2y – 3x + 6 = 0, rewrite it as 2y + 6 = 3x.

You can say that 2y + 6 is a multiple of 3 (=3x). So all values which must be integer must also satisfy this constraint. I'm just saying that, so it's easier to evaluate all the possible numbers (-6, -3, 0). If you plug in y = 0, x will be 2 and xy = 0 as the lowest possible value.

Hence, Answer Choice C is the one to go.
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Originally posted by reto on 20 May 2015, 14:19.
Last edited by reto on 21 May 2015, 14:20, edited 2 times in total.
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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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New post 20 May 2015, 22:43
4
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
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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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New post 21 May 2015, 04:39
2
reto wrote:

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=<y<=0

Ok, so far so good, we're half way through. What about x?

Here's the formula: 2y – 3x + 6 = 0, rewrite it as 2y + 6 = 3x. Now it's all about finding extreme values by combining with -6=<y<=0! If you play with the numbers and plug in y = -6, x will be -2 and the product would be 12. That's not what we want... if you plug in -2 for y, x will be 2/3 and the product xy is -2*2/3 ... since with all trials i don't get any of the numbers in the question steam, I assume the ANSWER is E.


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
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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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New post 21 May 2015, 06:23
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
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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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New post 22 May 2015, 04:28
1
-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.
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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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New post 22 May 2015, 08:03
csirishac wrote:
-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.



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.

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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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New post 10 Nov 2015, 02:50
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?
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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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New post 10 Nov 2015, 03:07
3
kanigmat011 wrote:
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?


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.
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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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New post 03 Dec 2015, 10:38
EgmatQuantExpert wrote:
reto wrote:

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=<y<=0

Ok, so far so good, we're half way through. What about x?

Here's the formula: 2y – 3x + 6 = 0, rewrite it as 2y + 6 = 3x. Now it's all about finding extreme values by combining with -6=<y<=0! If you play with the numbers and plug in y = -6, x will be -2 and the product would be 12. That's not what we want... if you plug in -2 for y, x will be 2/3 and the product xy is -2*2/3 ... since with all trials i don't get any of the numbers in the question steam, I assume the ANSWER is E.


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


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.
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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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New post 07 Dec 2015, 20:38
UJs wrote:
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



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.
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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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New post 07 Dec 2015, 21:17
robu wrote:
UJs wrote:
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



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.


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
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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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New post 28 Oct 2016, 07:11
EgmatQuantExpert wrote:
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


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.
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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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New post 26 Jul 2017, 05:33
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!
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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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New post 26 Jul 2017, 05:46
Madhavi1990 wrote:
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!


Yes Indeed....This approach is correct :)
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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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New post 09 Dec 2017, 09:33
If you visualize |y+3|<=3 on the number line you will get to know that y can take values in the range [-6,0]

Next express the give relation in terms of x i.e. y= (2/3)*(y-3).
and since y is in the range of [-6,0] 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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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,   [#permalink] 20 Jan 2019, 05:37
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