Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 20 Oct 2013
Posts: 50

Re: Graphic approach to problems with inequalities
[#permalink]
Show Tags
22 May 2014, 23:12
Bunuel wrote: mainhoon wrote: Can someone solve this without the graphical approach? Bunuel? My solution from: toughinequalitychallange89225.html?hilit=walker%20graphic#p732298If (x/y)>2, is 3x+2y<18? (1) xy is less than 2 (2) yx is less than 2 \(\frac{x}{y}>2\) tells us that \(x\) and \(y\) are either both positive or both negative, which means that all points \((x,y)\) satisfying given inequality are either in I or III quadrant. When they are both negative (in III quadrant) inequality \(3x+2y<18\) is always true, so we should check only for I quadrant, or when both \(x\) and \(y\) are positive. In I quadrant, as \(x\) and \(y\) are both positive, we can rewrite \(\frac{x}{y}>2\) as \(x>2y>0\) (remember \(x>0\) and \(y>0\)). So basically question becomes: If \(x>0\) and \(y>0\) and \(x>2y>0\), is \(3x+2y<18\)? (1) \(xy<2\). Subtract inequalities \(x>2y\) and \(xy<2\) (we can do this as signs are in opposite direction) > \(x(xy)>2y2\) > \(y<2\). Now add inequalities \(xy<2\) and \(y<2\) (we can do this as signs are in the same direction) > \(xy+y<2+2\) > \(x<4\). We got \(y<2\) and \(x<4\). If we take maximum values \(x=4\) and \(y=2\) and substitute in \(3x+2y<18\), we'll get \(12+4=16<18\). Sufficient. (2) \(yx<2\) and \(x>2y\): \(x=3\) and \(y=1\) > \(3x+2y=11<18\) true. \(x=11\) and \(y=5\) > \(3x+2y=43<18\) false. Not sufficient. Answer: A. Hi Bunnel I understood the part that x & y are either both positive or both negative.... when both are positive we get x>2y.... then we solve on that basis... what happens to when both are negative???? we solved both statements based on this and also declared statement 1 sufficient. when both x & y are negative then we cannot have x>2y right? so where are we considering this...



Math Expert
Joined: 02 Sep 2009
Posts: 65014

Re: Graphic approach to problems with inequalities
[#permalink]
Show Tags
23 May 2014, 01:19
NGGMAT wrote: Bunuel wrote: mainhoon wrote: Can someone solve this without the graphical approach? Bunuel? My solution from: toughinequalitychallange89225.html?hilit=walker%20graphic#p732298If (x/y)>2, is 3x+2y<18? (1) xy is less than 2 (2) yx is less than 2 \(\frac{x}{y}>2\) tells us that \(x\) and \(y\) are either both positive or both negative, which means that all points \((x,y)\) satisfying given inequality are either in I or III quadrant. When they are both negative (in III quadrant) inequality \(3x+2y<18\) is always true, so we should check only for I quadrant, or when both \(x\) and \(y\) are positive.In I quadrant, as \(x\) and \(y\) are both positive, we can rewrite \(\frac{x}{y}>2\) as \(x>2y>0\) (remember \(x>0\) and \(y>0\)). So basically question becomes: If \(x>0\) and \(y>0\) and \(x>2y>0\), is \(3x+2y<18\)? (1) \(xy<2\). Subtract inequalities \(x>2y\) and \(xy<2\) (we can do this as signs are in opposite direction) > \(x(xy)>2y2\) > \(y<2\). Now add inequalities \(xy<2\) and \(y<2\) (we can do this as signs are in the same direction) > \(xy+y<2+2\) > \(x<4\). We got \(y<2\) and \(x<4\). If we take maximum values \(x=4\) and \(y=2\) and substitute in \(3x+2y<18\), we'll get \(12+4=16<18\). Sufficient. (2) \(yx<2\) and \(x>2y\): \(x=3\) and \(y=1\) > \(3x+2y=11<18\) true. \(x=11\) and \(y=5\) > \(3x+2y=43<18\) false. Not sufficient. Answer: A. Hi Bunnel I understood the part that x & y are either both positive or both negative.... when both are positive we get x>2y.... then we solve on that basis... what happens to when both are negative???? we solved both statements based on this and also declared statement 1 sufficient. when both x & y are negative then we cannot have x>2y right? so where are we considering this... When they are both negative (in III quadrant) inequality \(3x+2y<18\) is always true, so we should check only for I quadrant, or when both \(x\) and \(y\) are positive.
_________________



Manager
Joined: 17 Jul 2013
Posts: 68

Re: Graphic approach to problems with inequalities
[#permalink]
Show Tags
24 Jun 2014, 03:24
Can we use this technique to solve any of the in equality ques. My accuracy in such questions suffer. Please mention how I should approach in equalities.....



Math Expert
Joined: 02 Sep 2009
Posts: 65014

Re: Graphic approach to problems with inequalities
[#permalink]
Show Tags
24 Jun 2014, 05:00
GmatDestroyer2013 wrote: Can we use this technique to solve any of the in equality ques. My accuracy in such questions suffer. Please mention how I should approach in equalities..... Graphic approach is not a silver bullet for inequality questions. Yes, it's handy for some of them but certainly not for all of them. Hope the links below help to attack them more efficiently:
_________________



Manager
Joined: 22 Jul 2014
Posts: 120
Concentration: General Management, Finance
WE: Engineering (Energy and Utilities)

Re: Graphic approach to problems with inequalities
[#permalink]
Show Tags
11 Aug 2014, 01:31
Can you please tell me how do you know which region to shade? I mean , to the left? or right? above or below the x axis? How do you find out



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10634
Location: Pune, India

Re: Graphic approach to problems with inequalities
[#permalink]
Show Tags
11 Aug 2014, 21:17
alphonsa wrote: Can you please tell me how do you know which region to shade? I mean , to the left? or right? above or below the x axis? How do you find out So you draw the line showing the equation represented by the inequality. How do you decide which side of the line does the inequality represent. Usually, you can do that by plugging in (0, 0) in the equation. The point (0, 0) will lie on one side of the line. Put x = 0 and y = 0 in your inequality. If it holds, it means the inequality holds for point (0, 0) and hence will hold for that entire side of the line. So you shade the side where (0, 0) lies. If the inequality does not hold when you put (0, 0), it means (0, 0) is not a solution of the inequality and hence the inequality holds for the opposite side so you shade the opposite side. If the line passes through (0, 0) try any other point which obviously lies on one side of the line.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Manager
Joined: 26 Feb 2015
Posts: 108

Re: Graphic approach to problems with inequalities
[#permalink]
Show Tags
13 Apr 2015, 00:16
I don't understand how we can deduce from (2) that its not sufficient. What I see from the graph, is that yx=2 never pass the red area. How am I supposed to interpret that graph?
All the other graphs makes sense "Oh okay, don't have any line passing the red part = sufficient"  but the last figure doesn't go through red zone either, yet still insufficient?
Am I supposed to realize that the lines can't pass through that white area just above "TRUE" either?



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10634
Location: Pune, India

Re: Graphic approach to problems with inequalities
[#permalink]
Show Tags
13 Apr 2015, 20:54
erikvm wrote: I don't understand how we can deduce from (2) that its not sufficient. What I see from the graph, is that yx=2 never pass the red area. How am I supposed to interpret that graph?
All the other graphs makes sense "Oh okay, don't have any line passing the red part = sufficient"  but the last figure doesn't go through red zone either, yet still insufficient?
Am I supposed to realize that the lines can't pass through that white area just above "TRUE" either? When the thread gets to the 6th page with, presumably, multiple solutions, you need to quote the post to which you are referring or give its link.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Manager
Joined: 26 Feb 2015
Posts: 108

Re: Graphic approach to problems with inequalities
[#permalink]
Show Tags
14 Apr 2015, 00:29
VeritasPrepKarishma wrote: erikvm wrote: I don't understand how we can deduce from (2) that its not sufficient. What I see from the graph, is that yx=2 never pass the red area. How am I supposed to interpret that graph?
All the other graphs makes sense "Oh okay, don't have any line passing the red part = sufficient"  but the last figure doesn't go through red zone either, yet still insufficient?
Am I supposed to realize that the lines can't pass through that white area just above "TRUE" either? When the thread gets to the 6th page with, presumably, multiple solutions, you need to quote the post to which you are referring or give its link. Figure 5. Original post. The line yx=2 never goes through the red zone  yet "insufficient"



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10634
Location: Pune, India

Re: Graphic approach to problems with inequalities
[#permalink]
Show Tags
14 Apr 2015, 21:14
erikvm wrote: VeritasPrepKarishma wrote: erikvm wrote: I don't understand how we can deduce from (2) that its not sufficient. What I see from the graph, is that yx=2 never pass the red area. How am I supposed to interpret that graph?
All the other graphs makes sense "Oh okay, don't have any line passing the red part = sufficient"  but the last figure doesn't go through red zone either, yet still insufficient?
Am I supposed to realize that the lines can't pass through that white area just above "TRUE" either? When the thread gets to the 6th page with, presumably, multiple solutions, you need to quote the post to which you are referring or give its link. Figure 5. Original post. The line yx=2 never goes through the red zone  yet "insufficient" It has nothing to do with whether it passes from the red region or not. You are given that y  x < 2. When you draw y  x = 2 in figure 5, you get a line. But what is y  x < 2? It is the area to the right of the line y  x = 2. You can find this by putting in (0, 0) in the inequality y  x < 2. You get 0  0 < 2 that is 0 < 2 which is true. Since (0, 0) lies to the right of the line and satisfies the inequality, it means the right of the line is y  x < 2 and the left of the line is y  x > 2. So you are given that points to the right of y  x = 2 are feasible. Look at all the points to the right of y  x= 2 and that satisfy x/y > 2 (i.e. the green and red regions). Note that of all the points to the right of y  x = 2, some lie in the green region and some lie in the red region (ignoring those that lie in neither region). The ones that lie in green region are those for which 3x+2y < 18. For those that lie in the red region, 3x + 2y > 18. So can we say whether 3x + 2y is less than 18? No. Because some points that satisfy y  x > 2 lie in the reg region and some lie in the green region. When will an inequality be sufficient. It will be sufficient when the region it points to has points lying only in one region  either red or green. Mind you, using graphing for inequalities isn't very easy at first. Only once you know instinctively what you are doing will you enjoy it. Here are three posts that build up on these concepts. http://www.veritasprep.com/blog/2010/12 ... hegraphs/http://www.veritasprep.com/blog/2010/12 ... spartii/http://www.veritasprep.com/blog/2011/01 ... partiii/
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Intern
Joined: 11 Aug 2014
Posts: 9
Location: United States
Concentration: International Business, Entrepreneurship
GPA: 3.9

Re: Graphic approach to problems with inequalities
[#permalink]
Show Tags
18 May 2016, 23:20
Bunuel / Chetan2u / Karishma : Your solution to this problem please ?
Thanks Regards



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10634
Location: Pune, India

Re: Graphic approach to problems with inequalities
[#permalink]
Show Tags
19 May 2016, 21:06
parvgugnani wrote: Bunuel / Chetan2u / Karishma : Your solution to this problem please ?
Thanks Regards You can also consider number line. 1. x  y < 2 ________________________y __ __ x____________x cannot be 2 units or more to the right of y. It is either closer or to the left of y (in the green region) But x/y > 2 so x is more than twice of y. If y is positive, y must be less than 2. So x will be less than 4 (cannot be 2 or more away from y). y can take any negative value. x would be to the left of y. When x is to the left of y, it can take any value. (If say y is 1, x could be 2.5; If y is 1, x could be 2.5) Is 3x + 2y < 18? x is less than 4 and y is less than 2 so 3x + 2y is less than 3*4 + 2*2 i.e. 16. Sufficient. 2. y  x < 2 _________________________x __ __ y ___________y cannot be 2 units or more to the right of x. It is either closer or to the left of x. (in the green region) But x/y > 2 so x is more than twice of y. If y is positive, any value of y is acceptable. y will be to the left of x. Is 3x + 2y < 18? We immediately note that since x and y can take large positive values or small ones, 3x + 2y may or may not be less than 18. Not sufficient. Answer (A)
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Intern
Joined: 28 Nov 2017
Posts: 16
Location: India
Concentration: General Management
GPA: 3.16
WE: Engineering (Energy and Utilities)

Re: Graphic approach to problems with inequalities
[#permalink]
Show Tags
26 Apr 2018, 09:36
Bunuel wrote: If \(\frac{x}{y} > 2\), is \(3x + 2y < 18\)?\(\frac{x}{y}>2\) tells us that \(x\) and \(y\) are either both positive or both negative, which means that all points \((x,y)\) satisfying given inequality are either in I or III quadrant. When they are both negative (in III quadrant) inequality \(3x+2y<18\) is always true, so we should check only for I quadrant, or when both \(x\) and \(y\) are positive. In I quadrant, as \(x\) and \(y\) are both positive, we can rewrite \(\frac{x}{y}>2\) as \(x>2y>0\) (remember \(x>0\) and \(y>0\)). So basically question becomes: If \(x>0\) and \(y>0\) and \(x>2y>0\), is \(3x+2y<18\)? (1) \(xy<2\). Subtract inequalities \(x>2y\) and \(xy<2\) (we can do this as signs are in opposite direction) > \(x(xy)>2y2\) > \(y<2\). Now add inequalities \(xy<2\) and \(y<2\) (we can do this as signs are in the same direction) > \(xy+y<2+2\) > \(x<4\). We got \(y<2\) and \(x<4\). If we take maximum values \(x=4\) and \(y=2\) and substitute in \(3x+2y<18\), we'll get \(12+4=16<18\). Sufficient. (2) \(yx<2\) and \(x>2y\): \(x=3\) and \(y=1\) > \(3x+2y=11<18\) true. \(x=11\) and \(y=5\) > \(3x+2y=43<18\) false. Not sufficient. Answer: A.THIS QUESTION IS DISCUSSED HERE. Bunuel I have one doubt. if we see the algebraic approach, the the first condition gives us the solution as y<2 and x<4. However, if i see the graphical approach, these values doesn't satisfy the graphical area satisfied by original condition and condition1. Clarify where am I going wrong?



Math Expert
Joined: 02 Sep 2009
Posts: 65014

Re: Graphic approach to problems with inequalities
[#permalink]
Show Tags
26 Apr 2018, 12:12
gmatacer40 wrote: Bunuel wrote: If \(\frac{x}{y} > 2\), is \(3x + 2y < 18\)?\(\frac{x}{y}>2\) tells us that \(x\) and \(y\) are either both positive or both negative, which means that all points \((x,y)\) satisfying given inequality are either in I or III quadrant. When they are both negative (in III quadrant) inequality \(3x+2y<18\) is always true, so we should check only for I quadrant, or when both \(x\) and \(y\) are positive. In I quadrant, as \(x\) and \(y\) are both positive, we can rewrite \(\frac{x}{y}>2\) as \(x>2y>0\) (remember \(x>0\) and \(y>0\)). So basically question becomes: If \(x>0\) and \(y>0\) and \(x>2y>0\), is \(3x+2y<18\)? (1) \(xy<2\). Subtract inequalities \(x>2y\) and \(xy<2\) (we can do this as signs are in opposite direction) > \(x(xy)>2y2\) > \(y<2\). Now add inequalities \(xy<2\) and \(y<2\) (we can do this as signs are in the same direction) > \(xy+y<2+2\) > \(x<4\). We got \(y<2\) and \(x<4\). If we take maximum values \(x=4\) and \(y=2\) and substitute in \(3x+2y<18\), we'll get \(12+4=16<18\). Sufficient. (2) \(yx<2\) and \(x>2y\): \(x=3\) and \(y=1\) > \(3x+2y=11<18\) true. \(x=11\) and \(y=5\) > \(3x+2y=43<18\) false. Not sufficient. Answer: A.THIS QUESTION IS DISCUSSED HERE. Bunuel I have one doubt. if we see the algebraic approach, the the first condition gives us the solution as y<2 and x<4. However, if i see the graphical approach, these values doesn't satisfy the graphical area satisfied by original condition and condition1. Clarify where am I going wrong? Check the graph below: Blue region satisfies the following conditions: \(x>0\) and \(y>0\) and \(x>2y>0\) \(xy<2\). As you can see for that region \(y<2\) and \(x<4\) holds true. Attachment:
Graph+.png [ 12.12 KiB  Viewed 854 times ]
_________________



Intern
Joined: 28 Nov 2017
Posts: 16
Location: India
Concentration: General Management
GPA: 3.16
WE: Engineering (Energy and Utilities)

Re: Graphic approach to problems with inequalities
[#permalink]
Show Tags
26 Apr 2018, 19:07
Bunuel wrote: gmatacer40 wrote: Bunuel wrote: If \(\frac{x}{y} > 2\), is \(3x + 2y < 18\)?\(\frac{x}{y}>2\) tells us that \(x\) and \(y\) are either both positive or both negative, which means that all points \((x,y)\) satisfying given inequality are either in I or III quadrant. When they are both negative (in III quadrant) inequality \(3x+2y<18\) is always true, so we should check only for I quadrant, or when both \(x\) and \(y\) are positive. In I quadrant, as \(x\) and \(y\) are both positive, we can rewrite \(\frac{x}{y}>2\) as \(x>2y>0\) (remember \(x>0\) and \(y>0\)). So basically question becomes: If \(x>0\) and \(y>0\) and \(x>2y>0\), is \(3x+2y<18\)? (1) \(xy<2\). Subtract inequalities \(x>2y\) and \(xy<2\) (we can do this as signs are in opposite direction) > \(x(xy)>2y2\) > \(y<2\). Now add inequalities \(xy<2\) and \(y<2\) (we can do this as signs are in the same direction) > \(xy+y<2+2\) > \(x<4\). We got \(y<2\) and \(x<4\). If we take maximum values \(x=4\) and \(y=2\) and substitute in \(3x+2y<18\), we'll get \(12+4=16<18\). Sufficient. (2) \(yx<2\) and \(x>2y\): \(x=3\) and \(y=1\) > \(3x+2y=11<18\) true. \(x=11\) and \(y=5\) > \(3x+2y=43<18\) false. Not sufficient. Answer: A.THIS QUESTION IS DISCUSSED HERE. Bunuel I have one doubt. if we see the algebraic approach, the the first condition gives us the solution as y<2 and x<4. However, if i see the graphical approach, these values doesn't satisfy the graphical area satisfied by original condition and condition1. Clarify where am I going wrong? Check the graph below: Blue region satisfies the following conditions: \(x>0\) and \(y>0\) and \(x>2y>0\) \(xy<2\). As you can see for that region \(y<2\) and \(x<4\) holds true. BunuelBunuel, what if we select values x=3.9 and y=0.1. It satisfies x>0 y>0 x<4 and y<2. However, I observe that this point doesn't come in graphical solution.please explain. Posted from my mobile device



Math Expert
Joined: 02 Sep 2009
Posts: 65014

Re: Graphic approach to problems with inequalities
[#permalink]
Show Tags
26 Apr 2018, 22:28
gmatacer40 wrote: Bunuel wrote: gmatacer40 wrote: If \(\frac{x}{y} > 2\), is \(3x + 2y < 18\)?\(\frac{x}{y}>2\) tells us that \(x\) and \(y\) are either both positive or both negative, which means that all points \((x,y)\) satisfying given inequality are either in I or III quadrant. When they are both negative (in III quadrant) inequality \(3x+2y<18\) is always true, so we should check only for I quadrant, or when both \(x\) and \(y\) are positive. In I quadrant, as \(x\) and \(y\) are both positive, we can rewrite \(\frac{x}{y}>2\) as \(x>2y>0\) (remember \(x>0\) and \(y>0\)). So basically question becomes: If \(x>0\) and \(y>0\) and \(x>2y>0\), is \(3x+2y<18\)? (1) \(xy<2\). Subtract inequalities \(x>2y\) and \(xy<2\) (we can do this as signs are in opposite direction) > \(x(xy)>2y2\) > \(y<2\). Now add inequalities \(xy<2\) and \(y<2\) (we can do this as signs are in the same direction) > \(xy+y<2+2\) > \(x<4\). We got \(y<2\) and \(x<4\). If we take maximum values \(x=4\) and \(y=2\) and substitute in \(3x+2y<18\), we'll get \(12+4=16<18\). Sufficient. (2) \(yx<2\) and \(x>2y\): \(x=3\) and \(y=1\) > \(3x+2y=11<18\) true. \(x=11\) and \(y=5\) > \(3x+2y=43<18\) false. Not sufficient. Answer: A.THIS QUESTION IS DISCUSSED HERE. Bunuel I have one doubt. if we see the algebraic approach, the the first condition gives us the solution as y<2 and x<4. However, if i see the graphical approach, these values doesn't satisfy the graphical area satisfied by original condition and condition1. Clarify where am I going wrong? Check the graph below: Blue region satisfies the following conditions: \(x>0\) and \(y>0\) and \(x>2y>0\) \(xy<2\). As you can see for that region \(y<2\) and \(x<4\) holds true. BunuelBunuel, what if we select values x=3.9 and y=0.1. It satisfies x>0 y>0 x<4 and y<2. However, I observe that this point doesn't come in graphical solution.please explain. Posted from my mobile deviceI see what you mean. x < 4 and y < 2 does NOT mean that ALL points which have have xcoordinate less than 4 and ycoordinate less than 2 will satisfy \(x>0\), \(y>0\), \(x>2y>0\) and\(xy<2\). It meas that ALL points which satisfy \(x>0\), \(y>0\), \(x>2y>0\) and\(xy<2\) will have xcoordinate less than 4 and ycoordinate less than 2. Or in another way, ALL points in blue region have xcoordinate less than 4 and ycoordinate less than 2 but NOT all points which have xcoordinate less than 4 and ycoordinate less than 2 are in blue region. Do you see the difference? Does this make sense?
_________________



Intern
Joined: 28 Nov 2017
Posts: 16
Location: India
Concentration: General Management
GPA: 3.16
WE: Engineering (Energy and Utilities)

Re: Graphic approach to problems with inequalities
[#permalink]
Show Tags
26 Apr 2018, 23:55
gmatacer40 wrote: Bunuel wrote: If \(\frac{x}{y} > 2\), is \(3x + 2y < 18\)?\(\frac{x}{y}>2\) tells us that \(x\) and \(y\) are either both positive or both negative, which means that all points \((x,y)\) satisfying given inequality are either in I or III quadrant. When they are both negative (in III quadrant) inequality \(3x+2y<18\) is always true, so we should check only for I quadrant, or when both \(x\) and \(y\) are positive. In I quadrant, as \(x\) and \(y\) are both positive, we can rewrite \(\frac{x}{y}>2\) as \(x>2y>0\) (remember \(x>0\) and \(y>0\)). So basically question becomes: If \(x>0\) and \(y>0\) and \(x>2y>0\), is \(3x+2y<18\)? (1) \(xy<2\). Subtract inequalities \(x>2y\) and \(xy<2\) (we can do this as signs are in opposite direction) > \(x(xy)>2y2\) > \(y<2\). Now add inequalities \(xy<2\) and \(y<2\) (we can do this as signs are in the same direction) > \(xy+y<2+2\) > \(x<4\). We got \(y<2\) and \(x<4\). If we take maximum values \(x=4\) and \(y=2\) and substitute in \(3x+2y<18\), we'll get \(12+4=16<18\). Sufficient. (2) \(yx<2\) and \(x>2y\): \(x=3\) and \(y=1\) > \(3x+2y=11<18\) true. \(x=11\) and \(y=5\) > \(3x+2y=43<18\) false. Not sufficient. Answer: A.THIS QUESTION IS DISCUSSED HERE. Bunuel I have one doubt. if we see the algebraic approach, the the first condition gives us the solution as y<2 and x<4. However, if i see the graphical approach, these values doesn't satisfy the graphical area satisfied by original condition and condition1. Clarify where am I going wrong? BunuelI see. So, as per our discussion, I conclude that algebraically method of solving has a limitation in the way that it provides superset of solution unlike graphical approach, which gives precisely the solution set only. Am I correct? However, it seems weird to me since I used to see inequality algebraic method in the other way. For example, if x>5 and y<2 that implies xy>3; the solution is so precise that every point greater than 3 satisfies the equation. So, still not very clear.



Math Expert
Joined: 02 Sep 2009
Posts: 65014

Re: Graphic approach to problems with inequalities
[#permalink]
Show Tags
27 Apr 2018, 02:52
gmatacer40 wrote: BunuelI see. So, as per our discussion, I conclude that algebraically method of solving has a limitation in the way that it provides superset of solution unlike graphical approach, which gives precisely the solution set only. Am I correct? However, it seems weird to me since I used to see inequality algebraic method in the other way. For example, if x>5 and y<2 that implies xy>3; the solution is so precise that every point greater than 3 satisfies the equation. So, still not very clear. Not entirely. The algebraic way, gives you precise ranges too. The point is that for (1) we not only have that x < 4 and y < 2, we also have the other constraints: \(x>0\), \(y>0\), \(x>2y>0\) and\(xy<2\). If you apply all of them you'll get the same domain as we got from the graphic approach, namely the blue region in my post above.
_________________



Intern
Joined: 29 Jun 2017
Posts: 19
Location: India
GPA: 3.48

Re: Graphic approach to problems with inequalities
[#permalink]
Show Tags
11 Nov 2018, 06:59
Hi, I have an alternate approach for the same. Please correct if Im wrong.
Given: x/y > 2 => Inference: a) x,y are both +ve or both are ve b) x > y
Now, (1) x  y < 2 if both are positive, then , let x = 7, y = 1 7  1 is not LT 2 if both are negative, then, let x = 7, y = 1 7  (1) => 6 < 2 (True) hence both x and y are negative therefore in the eq 3x + 2y => always < 0 => sufficient
(2) y  x < 2 if both positive, let x = 7, y = 1, 1  7 = 6 < 2 (True) if both negative, let x = 7, y = 1, 1  (7) => 7 not LT 2 Hence both are positive therefore, if positive values are put in the inequality 3x + 2y => clearly insufficient
Hence, answer is A.



NonHuman User
Joined: 09 Sep 2013
Posts: 15384

Re: Graphic approach to problems with inequalities
[#permalink]
Show Tags
14 Nov 2019, 09:55
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________




Re: Graphic approach to problems with inequalities
[#permalink]
14 Nov 2019, 09:55



Go to page
Previous
1 2 3
[ 60 posts ]

