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Bunuel, can you please explain the graphical method to solve this question?

Thanks

Dear nitin6305, I'm happy to help with this.

Idea #1: to graph an inequality, solve for slope-intercept form (i.e. y = mx + b form). When we do that Graph (1) becomes: y > 3x/2 + 1/2 Graph (2) becomes: y > 2x/3 + 1/3 We plot the exact lines (y = 3x/2 + 1/2 and y = 2x/3 + 1/3) to determine the boundaries of the regions. If the inequality is y > mx + b, then the region representing the inequality is above the line; if the inequality is y < mx + b, then the region representing the inequality is below the line.

inequalities with y = x.JPG [ 82.84 KiB | Viewed 5861 times ]

The darker green region, above the line y > 3x/2 + 1/23, is the region representing the first inequality. The yellow region, above the line y = 2x/3 + 1/3, is the region representing the second inequality. The bright spring green region is their overlap, the region satisfied by both inequalities. Notice that everything in that bright spring green region is above the solid green line, which is y = x. Points above the line y = x always have y > x.

Thanks for sharing the succinct graphical approach. I am having trouble understanding how the graph 2 gives a definite answer.

per the graph2 y > 2x/3 + 1/3 ------> the yellow region in the graph

I can see in the yellow region we have one small region where x>y and one bigger region where y>x. If that's true then we can't have a definite answer.

Please help, I think I have confused things here since the OA is B.

Thanks in advance !

Regards, SR

Dear solitaryreaper, I'm happy to help.

Statement #2 is not sufficient by itself. For most of the region 2x−3<3y−4, it's true that x < y.

For example, if (x, y) = (0, 5), then that is consistent with the statement #2 inequality, and it gives a "yes" answer to the prompt question, "is x < y?"

By contrast, if (x, y) = (9, 7), then this is also consistent with the statement #2 inequality, but it gives a "no" answer to the prompt question.

We can pick different values consistent with the statement #2 inequality that give different answers to the prompt question, so Statement #2 by itself is most certainly not sufficient.

Part of the problem is that whoever posted the question posted the WRONG OA! The correct OA for this question is (C), as the graphs I posted above show. I don't have the ability to made that edit, but perhaps Bunuel can make that change.

Does this answer your questions? Mike

Hi Mike,

The stem says that x and y are negative numbers, so the OA is correct.
_________________

Thanks for sharing the succinct graphical approach. I am having trouble understanding how the graph 2 gives a definite answer.

per the graph2 y > 2x/3 + 1/3 ------> the yellow region in the graph

I can see in the yellow region we have one small region where x>y and one bigger region where y>x. If that's true then we can't have a definite answer.

Please help, I think I have confused things here since the OA is B.

Thanks in advance !

Regards, SR

Dear solitaryreaper, I'm happy to help.

Statement #2 is not sufficient by itself. For most of the region 2x−3<3y−4, it's true that x < y.

For example, if (x, y) = (0, 5), then that is consistent with the statement #2 inequality, and it gives a "yes" answer to the prompt question, "is x < y?"

By contrast, if (x, y) = (9, 7), then this is also consistent with the statement #2 inequality, but it gives a "no" answer to the prompt question.

We can pick different values consistent with the statement #2 inequality that give different answers to the prompt question, so Statement #2 by itself is most certainly not sufficient.

Part of the problem is that whoever posted the question posted the WRONG OA! The correct OA for this question is (C), as the graphs I posted above show. I don't have the ability to made that edit, but perhaps Bunuel can make that change.

Does this answer your questions? Mike
_________________

Can you please explain this concept about evaluating the relevant region of an inequality? I tried reading your previous links but could not grasp it fully! ;(

Would really appreciate an explanation with graphs!! Thanks!

I am no expert but let me give it a try.

As per mikemcgarry 's graph above, y=x is a line passing through (0,0). All points on this line will be such that their y-coordinates will be = x-ccordinates. Thus examples points on this line will be (0,0),(1,1),(-10,-10) etc.

Remember that for this question, we need to only look in the 3rd quadrant (as both x and y are <0). ....(a)

Now take statement 1, 3x+4<2y+3. You can treat this as the linear equation, 3x+4=2y+3, giving you y=1.5x+0.5.

You now plot y=1.5x+0.5 and points following the inequality 3x+4<2y+3 or y>1.5x+0.5 will lie ABOVE the line y=1.5x+0.5 (additionally, had the inequality been y<1.5x+0.5, then you would have looked at all the points BELOW y=1.5x+0.5).

Additionally, see that y=1.5x+0.5 intersects y=x at (-1,-1) and this point is valid as per (a) above. Since there is a distinct (and valid) point of intersection between y=x and y=1.5x+0.5, you will have 2 scenarios with 1 scenario giving you a "yes" for y>x while for the other you will get a "no" for "y>x", making statement 1 not sufficient.

You can adopt the method above for statement 2 and see that as there are no points of intersection for y=x and y=0.67x+0.33 satisfying (a) above. Thus statement 2 will give a definite answer for "is y>x". Thus this statement is sufficient.

Re: If x and y are negative numbers, is x<y? [#permalink]

Show Tags

13 Aug 2013, 12:18

mikemcgarry wrote:

nitin6305 wrote:

If x and y are negative numbers, is x<y?

(1) 3x+4<2y+3

(2) 2x−3<3y−4

Bunuel, can you please explain the graphical method to solve this question?

Thanks

Dear nitin6305, I'm happy to help with this.

Idea #1: to graph an inequality, solve for slope-intercept form (i.e. y = mx + b form). When we do that Graph (1) becomes: y > 3x/2 + 1/2 Graph (2) becomes: y > 2x/3 + 1/3 We plot the exact lines (y = 3x/2 + 1/2 and y = 2x/3 + 1/3) to determine the boundaries of the regions. If the inequality is y > mx + b, then the region representing the inequality is above the line; if the inequality is y < mx + b, then the region representing the inequality is below the line.

The darker green region, above the line y > 3x/2 + 1/23, is the region representing the first inequality. The yellow region, above the line y = 2x/3 + 1/3, is the region representing the second inequality. The bright spring green region is their overlap, the region satisfied by both inequalities. Notice that everything in that bright spring green region is above the solid green line, which is y = x. Points above the line y = x always have y > x.

Re: If x and y are negative numbers, is x<y? [#permalink]

Show Tags

16 Sep 2014, 23:14

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).

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Re: If x and y are negative numbers, is x<y? [#permalink]

Show Tags

07 Oct 2015, 07:03

mikemcgarry wrote:

nitin6305 wrote:

If x and y are negative numbers, is x<y?

(1) 3x+4<2y+3

(2) 2x−3<3y−4

Bunuel, can you please explain the graphical method to solve this question?

Thanks

Dear nitin6305, I'm happy to help with this.

Idea #1: to graph an inequality, solve for slope-intercept form (i.e. y = mx + b form). When we do that Graph (1) becomes: y > 3x/2 + 1/2 Graph (2) becomes: y > 2x/3 + 1/3 We plot the exact lines (y = 3x/2 + 1/2 and y = 2x/3 + 1/3) to determine the boundaries of the regions. If the inequality is y > mx + b, then the region representing the inequality is above the line; if the inequality is y < mx + b, then the region representing the inequality is below the line.

The darker green region, above the line y > 3x/2 + 1/23, is the region representing the first inequality. The yellow region, above the line y = 2x/3 + 1/3, is the region representing the second inequality. The bright spring green region is their overlap, the region satisfied by both inequalities. Notice that everything in that bright spring green region is above the solid green line, which is y = x. Points above the line y = x always have y > x.

Thanks for sharing the succinct graphical approach. I am having trouble understanding how the graph 2 gives a definite answer.

per the graph2 y > 2x/3 + 1/3 ------> the yellow region in the graph

I can see in the yellow region we have one small region where x>y and one bigger region where y>x. If that's true then we can't have a definite answer.

Please help, I think I have confused things here since the OA is B.

Re: If x and y are negative numbers, is x<y? [#permalink]

Show Tags

07 Oct 2015, 22:07

statement 1. 3x+1<2y,now both no's are negative ,x<y/(3/2)-1/3=y/1.5-1/3 (y/1.5 is increased as negative no,e.g, -1.5<-1(-1.5/1.5) and then -1/3 added,we can't say x<y .

Statement 2. 2x<3y-1 or x<(y x 1.5) -1/2, negative number multiplied by a positive will give a lesser number then -1 added ..again decreased

when x<z where z is definitely lesser than y so x<y ,2 is enough

Re: If x and y are negative numbers, is x<y? [#permalink]

Show Tags

10 Oct 2015, 06:16

Bunuel wrote:

mikemcgarry wrote:

solitaryreaper wrote:

Hi mikemcgarry

Thanks for sharing the succinct graphical approach. I am having trouble understanding how the graph 2 gives a definite answer.

per the graph2 y > 2x/3 + 1/3 ------> the yellow region in the graph

I can see in the yellow region we have one small region where x>y and one bigger region where y>x. If that's true then we can't have a definite answer.

Please help, I think I have confused things here since the OA is B.

Thanks in advance !

Regards, SR

Dear solitaryreaper, I'm happy to help.

Statement #2 is not sufficient by itself. For most of the region 2x−3<3y−4, it's true that x < y.

For example, if (x, y) = (0, 5), then that is consistent with the statement #2 inequality, and it gives a "yes" answer to the prompt question, "is x < y?"

By contrast, if (x, y) = (9, 7), then this is also consistent with the statement #2 inequality, but it gives a "no" answer to the prompt question.

We can pick different values consistent with the statement #2 inequality that give different answers to the prompt question, so Statement #2 by itself is most certainly not sufficient.

Part of the problem is that whoever posted the question posted the WRONG OA! The correct OA for this question is (C), as the graphs I posted above show. I don't have the ability to made that edit, but perhaps Bunuel can make that change.

Does this answer your questions? Mike

Hi Mike,

The stem says that x and y are negative numbers, so the OA is correct.

Thanks a lot mikemcgarry for explaining the whole thing.

Kudos to Bunuel for pointing out our mistake. In wake of x<0 (given in question stem) , inequality y>x holds true. Hence statement 2 is sufficient. OA should be B.

If x and y are negative numbers, is x<y? [#permalink]

Show Tags

10 Oct 2015, 08:37

mikemcgarry wrote:

nitin6305 wrote:

If x and y are negative numbers, is x<y?

(1) 3x+4<2y+3

(2) 2x−3<3y−4

Bunuel, can you please explain the graphical method to solve this question?

Thanks

Dear nitin6305, I'm happy to help with this.

Idea #1: to graph an inequality, solve for slope-intercept form (i.e. y = mx + b form). When we do that Graph (1) becomes: y > 3x/2 + 1/2 Graph (2) becomes: y > 2x/3 + 1/3 We plot the exact lines (y = 3x/2 + 1/2 and y = 2x/3 + 1/3) to determine the boundaries of the regions. If the inequality is y > mx + b, then the region representing the inequality is above the line; if the inequality is y < mx + b, then the region representing the inequality is below the line.

The darker green region, above the line y > 3x/2 + 1/23, is the region representing the first inequality. The yellow region, above the line y = 2x/3 + 1/3, is the region representing the second inequality. The bright spring green region is their overlap, the region satisfied by both inequalities. Notice that everything in that bright spring green region is above the solid green line, which is y = x. Points above the line y = x always have y > x.

Does all this make sense? Mike

Hi Mike,

Can you please explain this concept about evaluating the relevant region of an inequality? I tried reading your previous links but could not grasp it fully! ;(

Would really appreciate an explanation with graphs!! Thanks!

Re: If x and y are negative numbers, is x<y? [#permalink]

Show Tags

10 Oct 2015, 09:43

Engr2012 wrote:

longfellow wrote:

Hi Mike,

Can you please explain this concept about evaluating the relevant region of an inequality? I tried reading your previous links but could not grasp it fully! ;(

Would really appreciate an explanation with graphs!! Thanks!

I am no expert but let me give it a try.

As per mikemcgarry 's graph above, y=x is a line passing through (0,0). All points on this line will be such that their y-coordinates will be = x-ccordinates. Thus examples points on this line will be (0,0),(1,1),(-10,-10) etc.

Remember that for this question, we need to only look in the 3rd quadrant (as both x and y are <0). ....(a)

Now take statement 1, 3x+4<2y+3. You can treat this as the linear equation, 3x+4=2y+3, giving you y=1.5x+0.5.

You now plot y=1.5x+0.5 and points following the inequality 3x+4<2y+3 or y>1.5x+0.5 will lie ABOVE the line y=1.5x+0.5 (additionally, had the inequality been y<1.5x+0.5, then you would have looked at all the points BELOW y=1.5x+0.5).

Additionally, see that y=1.5x+0.5 intersects y=x at (-1,-1) and this point is valid as per (a) above. Since there is a distinct (and valid) point of intersection between y=x and y=1.5x+0.5, you will have 2 scenarios with 1 scenario giving you a "yes" for y>x while for the other you will get a "no" for "y>x", making statement 1 not sufficient.

You can adopt the method above for statement 2 and see that as there are no points of intersection for y=x and y=0.67x+0.33 satisfying (a) above. Thus statement 2 will give a definite answer for "is y>x". Thus this statement is sufficient.

Re: If x and y are negative numbers, is x<y? [#permalink]

Show Tags

02 Dec 2016, 06:01

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.
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