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In the xy plane the area of the region bounded by the [#permalink]
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In the xy plane, the area of the region bounded by the graph of x+y + xy = 4 is A. 8 B 12 C. 16 D. 20 E. 24 Need help in solving equations involving Mod...... help?
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Last edited by Bunuel on 15 Feb 2012, 00:29, edited 2 times in total.
Edited the question and added the OA



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Re: graphs_Modulus....Help [#permalink]
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08 Nov 2009, 15:27
Bunuel wrote: papillon86 wrote: In xy plane, the area of the region bounded by the graph of x+y + xy = 4 is
a) 8 b) 12 c) 16 d) 20
Need help in solving equations involving Mod...... help? I've never seen such kind of question in GMAT before. 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 The area bounded by 4 graphs x=2, x=2, y=2, y=2 will be square with the side of 4 so the area will be 4*4=16. Answer: C Why cant we consider (4,0) and (0,4) as points on graph ? then area would be different... , right?
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08 Nov 2009, 15:39
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.
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Re: graphs_Modulus....Help [#permalink]
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08 Nov 2009, 16:58
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. 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
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Re: graphs_Modulus....Help [#permalink]
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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.
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Re: graphs_Modulus....Help [#permalink]
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08 Nov 2009, 20:23
Thanks Bunuel , once again wonderful explanation +1 Kudos.. have a good day...
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Re: graphs_Modulus....Help [#permalink]
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20 May 2011, 12:39
Bunuel wrote: x=2 x=2 y=2 y=2
This lines will make a square with the side 4, hence area 4*4=16.
i am still now able to follow if the area that is formed with these lines is as per fig 1 or fig 2
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Re: graphs_Modulus....Help [#permalink]
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20 May 2011, 17:31
Thats same as what you see in fig 2 . agdimple333 wrote: Bunuel wrote: x=2 x=2 y=2 y=2
This lines will make a square with the side 4, hence area 4*4=16.
i am still now able to follow if the area that is formed with these lines is as per fig 1 or fig 2



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Re: graphs_Modulus....Help [#permalink]
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20 May 2011, 17:33
solving the inequality we have the following as solutions
x=2 y=2 y=2 x=2
drawing this in a graph, we can observe that it forms a square with side length of 4.
Hence the area is 4*4 = 16
Answer is C.



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Re: graphs_Modulus....Help [#permalink]
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20 May 2011, 21:50
so check for ++, +, + and  giving x=22 and y=22 hence a square with 4*4 area = 16
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Re: graphs_Modulus....Help [#permalink]
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xy = xy if xy > 0 xy = (xy) if xy < 0 x+y > 0 => x > y then x !> y x+y + x  y = 4 x = 2 x  y + x  y = 4 (if x < y, then x !< y) y = 2 x + y x + y = 4 => y = 2 xy + x  y = 4 => y = 2 Answer  C
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Re: graphs_Modulus....Help [#permalink]
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22 May 2011, 07:21
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



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Re: graphs_Modulus....Help [#permalink]
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22 May 2011, 08:38
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: 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: graphs_Modulus....Help [#permalink]
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27 Feb 2012, 23:42
devinawilliam83 wrote: 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. Yes, it can be done by graphing. x+y + xy = 4 can expand in four different wasy: 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 So you can draw all these four lines x=2, x=2, y=2, y=2 to get a square with the side of 4: Attachment:
Square.gif [ 1.86 KiB  Viewed 16164 times ]
See more examples here: m065absolutevalue108191.htmlgraphsmodulushelp86549.htmlm06q572817.htmlifequationenclosesacertainregion110053.htmlHope it helps.
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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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eaakbari wrote: 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
Any absolute values such as x = 5 could mean that x = 5 or x = 5. Derive both () and (+) possibilities. For the problem: x+y + xy = 4 We could derive two possibilities for x+y could be (x+y) and (x+y) We could derive two possibilities for xy could be (xy) and (xy) This is the reason why we have 4 derived equations. (x+y) + (xy) = 4 (x+y)  (xy) = 4 (x+y) + (xy) = 4 (x+y)  (xy) = 4 Just simplify those... If you want more practice on this question type: http://burnoutorbreathe.blogspot.com/2012/12/absolutevaluessolvingforareaof.html
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