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Re: Absolute Values [#permalink]
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15 Feb 2012, 00:25



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Re: graphs_Modulus....Help [#permalink]
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27 Feb 2012, 23:33
Hi, Can this be solved by graphing. If yes .. how do we graph the equation with 2 mod parts VeritasPrepKarishma wrote: VinuPriyaN wrote: Given xy + x+y = 4
I don't understand why can't xy and x+y be 1 and 3 instead of 2 and 2! (which again equals 4)
Can any one please explain this to me?
Thanks & Regards, Vinu Look at the solution given by Bunuel above. When you solve it, you get four equations. One of them is x = 2 which means that x = 2 and y can take any value. If y = 1, xy = 1 and x+y = 3. For different values of y, xy and x+y will get different values. We are not discounting any of them.



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Re: In the xy plane the area of the region bounded by the [#permalink]
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05 Dec 2012, 23:17
(1) derive all equations from x+y + xy = 4 x+y+xy =4 ==> x=2 x+yx+y =4 ==> y=2 xy+xy =4 ==> y=2 xyx+y =4 ==> x=2 (2) Plot your four lines (3) Notice you have formed a square region bounded by x=2, y=2, y=2 and x=2 lines (4) Area = 4*4 = 16 Answer: C For more detailed solutions for similar question types:
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Re: graphs_Modulus....Help [#permalink]
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12 Dec 2012, 09:06
Quote: OK there can be 4 cases:
x+y + xy = 4
A. x+y+xy = 4 > x=2 B. x+yx+y = 4 > y=2 C. xy +xy= 4 > y=2 D. xyx+y=4 > x=2
Bunuel, Would appreciate it, if you could thoroughly explain the above. Thanks.
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Re: graphs_Modulus....Help [#permalink]
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13 Dec 2012, 04:26
Bunuel wrote: srini123 wrote: Thanks Bunuel, I used similar method for a similar question and I got wrong answer the question was
what is the area bounded by graph\(x/2 + y/2 = 5\)?
I got hunderd since x=10 x=10 y=10 y=10
isnt the area 400 ? the answer given was 200, please explain I think this one is different. \(\frac{x}{2} + \frac{y}{2} = 5\) After solving you'll get equation of four lines: \(y=10x\) \(y=10+x\) \(y=10x\) \(y=x10\) These four lines will also make a square, BUT in this case the diagonal will be 20 so the \(Area=\frac{20*20}{2}=200\). Or the \(Side= \sqrt{200}\), area=200. If you draw these four lines you'll see that the figure (square) which is bounded by them is turned by 90 degrees and has a center at the origin. So the side will not be 20. Also you made a mistake in solving equation. The red part is not correct. You should have the equations written above. In our original question when we were solving the equation x+y + xy = 4 each time x or y were cancelling out so we get equations of a type x=some value twice and y=some value twice. And these equations give the lines which are parallel to the Y or X axis respectively so the figure bounded by them is a "horizontal" square (in your question it's "diagonal" square). Hope it's clear. Hii Bunuel. What is the best approach of finding the points of intersection in order to make the square.
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Re: graphs_Modulus....Help [#permalink]
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13 Dec 2012, 04:30
Marcab wrote: Bunuel wrote: srini123 wrote: Thanks Bunuel, I used similar method for a similar question and I got wrong answer the question was
what is the area bounded by graph\(x/2 + y/2 = 5\)?
I got hunderd since x=10 x=10 y=10 y=10
isnt the area 400 ? the answer given was 200, please explain I think this one is different. \(\frac{x}{2} + \frac{y}{2} = 5\) After solving you'll get equation of four lines: \(y=10x\) \(y=10+x\) \(y=10x\) \(y=x10\) These four lines will also make a square, BUT in this case the diagonal will be 20 so the \(Area=\frac{20*20}{2}=200\). Or the \(Side= \sqrt{200}\), area=200. If you draw these four lines you'll see that the figure (square) which is bounded by them is turned by 90 degrees and has a center at the origin. So the side will not be 20. Also you made a mistake in solving equation. The red part is not correct. You should have the equations written above. In our original question when we were solving the equation x+y + xy = 4 each time x or y were cancelling out so we get equations of a type x=some value twice and y=some value twice. And these equations give the lines which are parallel to the Y or X axis respectively so the figure bounded by them is a "horizontal" square (in your question it's "diagonal" square). Hope it's clear. Hii Bunuel. What is the best approach of finding the points of intersection in order to make the square. I'd say substituting x=0 and y=0 in the equations of lines and making a drawing.
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Re: In the xy plane the area of the region bounded by the [#permalink]
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13 Dec 2012, 04:41



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Re: In the xy plane the area of the region bounded by the [#permalink]
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17 Jul 2013, 23:20
This is an interesting combo of absolute values, plotting lines in coordinate system and then finding the resulting figure's area. Thanks all for presenting the approach!



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Re: graphs_Modulus....Help [#permalink]
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21 Aug 2013, 21:51
Bunuel wrote: srini123 wrote: Why cant we consider (4,0) and (0,4) as points on graph ? then area would be different... , right? First of all we are not considering points separately, as we have XY plane and roots of equation will represent lines, we'll get the figure bounded by this 4 lines. The equations for the lines are: x=2 x=2 y=2 y=2 This lines will make a square with the side 4, hence area 4*4=16. Second: points (4,0) or (0,4) doesn't work for x+y + xy = 4. The side of the square can't be 4, instead its sqrt(8)
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Re: graphs_Modulus....Help [#permalink]
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22 Aug 2013, 03:21
honchos wrote: Bunuel wrote: srini123 wrote: Why cant we consider (4,0) and (0,4) as points on graph ? then area would be different... , right? First of all we are not considering points separately, as we have XY plane and roots of equation will represent lines, we'll get the figure bounded by this 4 lines. The equations for the lines are: x=2 x=2 y=2 y=2 This lines will make a square with the side 4, hence area 4*4=16. Second: points (4,0) or (0,4) doesn't work for x+y + xy = 4. The side of the square can't be 4, instead its sqrt(8)The side of the square IS 4: Attachment:
MSP39361d6ehgde6ie87a8800003827f7f92a367c60.gif [ 1.86 KiB  Viewed 1289 times ]
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Re: graphs_Modulus....Help [#permalink]
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22 Aug 2013, 05:36
Yes I realized it later in the day. Thanks for your help and support. The side of the square can't be 4, instead its sqrt(8)[/quote] The side of the square IS 4: Attachment: MSP39361d6ehgde6ie87a8800003827f7f92a367c60.gif [/quote]
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Re: In the xy plane the area of the region bounded by the [#permalink]
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27 Jan 2014, 03:00
Bunuel,
wouldn't it be sufficient to look at only two cases?
(x+y) + (xy) = 4 ==> x=2 (x+y)  (xy) = 4 ==> y=2
Which would give us 2*2 * 4 = 16?



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Re: In the xy plane the area of the region bounded by the [#permalink]
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Re: In the xy plane the area of the region bounded by the [#permalink]
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27 Jan 2014, 05:07
Bunuel wrote: unceldolan wrote: Bunuel,
wouldn't it be sufficient to look at only two cases?
(x+y) + (xy) = 4 ==> x=2 (x+y)  (xy) = 4 ==> y=2
Which would give us 2*2 * 4 = 16? What you mean by "sufficient"? x+y + xy = 4 gives FOUR equations, as explained on page 1. Hey Bunuel, I think I confused some things here. Just this morning, I did Chapter 9 of MGMAT Strategy Guide 2. Here it's stated that, if I have an equation with 2 absolute value expressions and one variable, I only need to set up 2 cases. Case A, when the absolute values have the same sign, Case B when the absolute values have different signs. But since here we got 2 variables and two absolute value expressions, I think I have to set up 4 equations. I just confused what I read a bit...sorry!



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Re: graphs_Modulus....Help [#permalink]
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14 Apr 2014, 04:09
HI Bunnel,
Have one doubt on this.
Similar kind of question is posted on following link
http://gmatclub.com/forum/m25q1976535.html
why here we are doing different then defined on above link.
on the above link we have square of side 20 then why we are not getting ans as 20*20 = 400



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Re: graphs_Modulus....Help [#permalink]
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14 Apr 2014, 04:40



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Re: graphs_Modulus....Help [#permalink]
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14 Apr 2014, 08:50
If we add all the lines in this way. then its diagonal is 4 so its side should be 2root2. In this case the area would be 8
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File comment: If we add all the lines in this way. then its diagonal is 4 so its side should be 2root2.
In this case the area would be 8
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Re: graphs_Modulus....Help [#permalink]
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14 Apr 2014, 20:47
pawankumargadiya wrote: If we add all the lines in this way. then its diagonal is 4 so its side should be 2root2.
In this case the area would be 8 The square is formed by lines x = 2, y = 2, x = 2 and y = 2 from the mod equation given in the question. You cannot turn it the way you did because you made the square smaller when you turned it. The side of the square needs to be 4 and diagonal needs to be 4*root2. Instead when you turned it, you made the diagonal 4 and side 4/root2 = 2*root2. That is incorrect. You can turn the square but it will not cut the axis at 2 or 2. It will cut the axis at 2*root2 or 2*root2. Then the square remains the same areawise.
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Re: graphs_Modulus....Help [#permalink]
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14 Apr 2014, 21:48
Hi karishma, Could you please elaborate more on this one . I am really having a hard time figuring this one out . I understood the first question i.e In xy plane, the area of the region bounded by the graph of x+y + xy = 4 . The area is 16. But in Second Question what is the area bounded by graphx/2 + y/2 = 5? Why was the graph drawn different from that of the previous Question. Why was diagonal considered for second question and Side of a square considered for first question. Help is appreciated . Thanks in advance VeritasPrepKarishma wrote: pawankumargadiya wrote: If we add all the lines in this way. then its diagonal is 4 so its side should be 2root2.
In this case the area would be 8 The square is formed by lines x = 2, y = 2, x = 2 and y = 2 from the mod equation given in the question. You cannot turn it the way you did because you made the square smaller when you turned it. The side of the square needs to be 4 and diagonal needs to be 4*root2. Instead when you turned it, you made the diagonal 4 and side 4/root2 = 2*root2. That is incorrect. You can turn the square but it will not cut the axis at 2 or 2. It will cut the axis at 2*root2 or 2*root2. Then the square remains the same areawise.




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