Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

In statement (1), you have said that we have these possibilities: (3r + 2 - s) = 0 OR (4r + 9 - s) = 0 OR both.

But, when we have something like this: (x + 3)(x+5) = 0

In this case, there are only two possibilities, right? (x+3)=0 OR (x+5)=0. Both cannot be zero at the same time because "x" represents a single and unique value. Please confirm.

Yes.

(3r + 2 - s) = 0 and (4r + 9 - s) = 0 can both be true, for r=-7 and s=-19.

But for (x+3)=0 and (x+5)=0 both cannot be true simultaneously. Either x=-3 or x=-5.

Re: In the XY plane, does the line with equation y=3x+2 [#permalink]

Show Tags

07 Jun 2014, 12:40

could we also think of it as:

the following two equations define two non-identical hyperbola. 1) (3r+2-s) (4r+9-s)=0 2) (4r-6-s) (3r+2-s)=0

each equation represents a set of points on different hyperbola since the question states that (r,s) is one point, it must be a single intersection of the two hyperbola which cannot be found without both equations.

therefore: you need both equations to define one point (r,s) and then you can easily solve for whether (r,s) satisfies the first equation y = 3x+2

In the XY plane, does the line with equation y=3x+2 [#permalink]

Show Tags

25 Nov 2014, 12:28

(Just so I get the whole picture) Say, instead of the bold parts in (3r+2-s)(4r+9-s)=0 and (4r-6-s)(3r+2-s)=0, there are 2 slightly unrelated terms like 5r-3-s and 2r+7-s.They would remain to be possibilities of the equation and E would be the right choice, right? Sorry if I am complicating things.

Last edited by deeuk on 26 Nov 2014, 11:36, edited 1 time in total.

(Just so I get the whole picture) Say, instead of the bold parts in (3r+2-s)(4r+9-s)=0 and (4r-6-s)(3r+2-s)=0, there were 2 slightly unrelated terms like 5r-3-s and 2r+7-s, they would remain to be possibilities of the equation and E would be the right choice, right? Sorry if I am complicating things.

Yes, in this case the answer would be E.
_________________

Re: In the XY plane, does the line with equation y=3x+2 [#permalink]

Show Tags

05 Aug 2016, 10:07

Substituting variables r and s for x and y is very smart. Then the question becomes a logic question.

(1) can contain the point or it could not (2) can contain the point or could not (just like statement 1)

Together at first it looks like it will not work out, however it has to have the point because the other two equations could not satisfy both of the equations. It would be 4r + 9 = 4r - 6 which would never work.

Re: In the XY plane, does the line with equation y=3x+2 [#permalink]

Show Tags

20 Aug 2016, 10:25

1

This post was BOOKMARKED

Guys, please say if I am wrong. If the question was found that 1) (3r+3-s) (4r+9-s)=0 2) (4r-6-s) (3r+3-s)=0 then we will conclud that the answer is C, do not contain since parallel

If the question was found that 1) (4r+2-s) (4r+9-s)=0 2) (4r-6-s) (4r+2-s)=0 then we will conclud that the answer is C, contain since intercept

Re: In the XY plane, does the line with equation y=3x+2 [#permalink]

Show Tags

30 Jan 2017, 18:24

(r,s) will lie on the line with equation y = 3x + 2, if in place of x and y respectively we put the coordinates of the point and the line equation is satisfied.

So , (r,s) will lie on y = 3x +2, if s=3r+2 or in other words:

3r-s+2 = 0.

From St. 1:

(3r + 2 - s)(4r + 9 - s) = 0 i.e.

either

(3r + 2 - s) = 0 or (4r + 9 - s) = 0.

Say (3r + 2 - s) = 0 -->St. 1 satisfied and (r, s) lies on given line.

Now say, (3r + 2 - s) not equal to zero, but (4r + 9 - s) = 0

Then St. 1 satisfied but (r, s) does not lie on given line.

So St. 1 is not sufficient.

Exactly same logic for St. 2.

St. 2 not sufficient.

Combining both statements.

case 1: (3r + 2 - s) is equal to zero

case 2: (3r + 2 - s) is not equal to zero

let's discuss case (2):

This means that

(4r + 9 - s) = 0 and also (4r - 6 - s) = 0

which is absurd. (4r-s = -9 as well as 6--> not possible)

Re: In the XY plane, does the line with equation y=3x+2 [#permalink]

Show Tags

01 Feb 2017, 07:33

Bunuel wrote:

In the XY plane, does the line with equation y=3x+2 contain the point (r,s)?

Line with equation \(y=3x+2\) contains the point \((r,s)\) means that when substituting \(r\) ans \(s\) in line equation: \(s=3r+2\) (or \(3r+2-s=0\)) holds true.

So basically we are asked to determine whether \(3r+2-s=0\) is true or not.

(1) \((3r+2-s)(4r+9-s)=0\) --> either \(3r+2-s=0\) OR \(4r+9-s=0\) OR both. Not sufficient.

(2) \((4r-6-s)(3r+2-s)=0\) --> either \(3r+2-s=0\) OR \(4r-6-s=0\) OR both. Not sufficient.

(1)+(2) Both \(4r+9-s=0\) and \(4r-6-s=0\) cannot be true (simultaneously), as \(4r-s\) can not equal to both -9 and 6, hence \(3r+2-s=0\) must be true. Sufficient.

Answer: C.

Hi,

This solution will not hold true if r and s are on a rotated axis on the plane i.e. r= xcosA + ysinA and s = -xsinA + ycosA for any A. In this case line representing 3r+2-s=0 will intersect 3x+2 = y at a single point, thus the line 3x+2 = y may or may not contain (r,s), hence, E.

In the XY plane, does the line with equation y=3x+2 contain the point (r,s)?

Line with equation \(y=3x+2\) contains the point \((r,s)\) means that when substituting \(r\) ans \(s\) in line equation: \(s=3r+2\) (or \(3r+2-s=0\)) holds true.

So basically we are asked to determine whether \(3r+2-s=0\) is true or not.

(1) \((3r+2-s)(4r+9-s)=0\) --> either \(3r+2-s=0\) OR \(4r+9-s=0\) OR both. Not sufficient.

(2) \((4r-6-s)(3r+2-s)=0\) --> either \(3r+2-s=0\) OR \(4r-6-s=0\) OR both. Not sufficient.

(1)+(2) Both \(4r+9-s=0\) and \(4r-6-s=0\) cannot be true (simultaneously), as \(4r-s\) can not equal to both -9 and 6, hence \(3r+2-s=0\) must be true. Sufficient.

Answer: C.

Hi,

This solution will not hold true if r and s are on a rotated axis on the plane i.e. r= xcosA + ysinA and s = -xsinA + ycosA for any A. In this case line representing 3r+2-s=0 will intersect 3x+2 = y at a single point, thus the line 3x+2 = y may or may not contain (r,s), hence, E.

Dear Kapidhwaj,

Not sure what are you trying to say but the only solution (3r+2-s)(4r+9-s)=0 and (4r-6-s)(3r+2-s)=0 has is 3r+2-s=0, which makes the answer C correct. Also, note that this is an official question and confirmed OA is C.
_________________

There’s something in Pacific North West that you cannot find anywhere else. The atmosphere and scenic nature are next to none, with mountains on one side and ocean on...

This month I got selected by Stanford GSB to be included in “Best & Brightest, Class of 2017” by Poets & Quants. Besides feeling honored for being part of...

Joe Navarro is an ex FBI agent who was a founding member of the FBI’s Behavioural Analysis Program. He was a body language expert who he used his ability to successfully...