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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. _________________

Saving was yesterday, heat up the gmatclub.forum's sentiment by spending KUDOS!

PS Please send me PM if I do not respond to your question within 24 hours.

Last edited by reto on 21 May 2015, 13:20, edited 2 times in total.

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.

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

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

Show Tags

21 May 2015, 05:23

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

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.

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.
_________________

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

Show Tags

03 Dec 2015, 09:38

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.

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

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.

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
_________________

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

Show Tags

28 Oct 2016, 06:11

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

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

Show Tags

26 Jul 2017, 04:33

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!

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

Show Tags

26 Jul 2017, 04:46

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!

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

Show Tags

09 Dec 2017, 08:33

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