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 350,000 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 the xy-plane, does the line with equation y=3x+2 contain the point (r,s)?

1, (3r+2-s)(4r + 9 -s) = 0

2, (4r -6 -s )(3r + 2 - s) = 0

Lets look at the question first. We have an equation giving relationship between x & y. And the question asks us if (r,s) is contained in y=3x+2

Usually when we want to see if if something is contained in a given line equation we have to sub those values and see if the equation holds good.
Ah so we should be looking for values of r & s in statements 1 and 2

Good now armed with this information lets look at statement 1
(3r+2-s)(4r + 9 -s) = 0
Great they already factored it for us so
3r+2-s = 0 --> 3r-s = -2 ------------Eqn(1)
4r+9-s = 0 ----> 4r-s = -9 ----------Eqn(2)

Solving Eqn (1) and Eqn(2) we get r= -7 and s= -19

Sub (r,s) in the question stem y=3x+2 ---> -19 = 3(-7) +2--> -19 = -19

Great Statement 1 is sufficient

Statement(2)
(4r -6 -s )(3r + 2 - s) = 0
This one is factored as well.
(4r -6 -s ) = 0 -------> 4r-s = 6 ------Eqn(3)
(3r+2-s) = 0 ------> 3r-s = -2 ------Eqn(4)

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

1, (3r+2-s)(4r + 9 -s) = 0

2, (4r -6 -s )(3r + 2 - s) = 0

Q, y=3x+2, intersects y axis at 2 and slope of 3 (imagine)

s=3r+2 (same line equation, definetely intersects), may or may not intersect
s=4r+9 (intersects y axis at 9 and more slope) so it has to intersect line in Q

From 2,
s = 4r-6(intersects y axis on -6 and slope of 4), may or may not intersect
s=3r+2(same line equation) definetely intersects

Does this make sense? Am i thinking in right direction?

My guess is B.
For the points (r,s) to lie on the line y=3x+2, the slope of r,s should be equal to the slope of y=3x+2 and since 2 provids us with this (s=3r+2) my guess is B.

My guess is B. For the points (r,s) to lie on the line y=3x+2, the slope of r,s should be equal to the slope of y=3x+2 and since 2 provids us with this (s=3r+2) my guess is B.

hmm, but the point (r,s) could be on a line parallell to y=3x+2?

My guess is B. For the points (r,s) to lie on the line y=3x+2, the slope of r,s should be equal to the slope of y=3x+2 and since 2 provids us with this (s=3r+2) my guess is B.

hmm, but the point (r,s) could be on a line parallell to y=3x+2?

I am not sure whether you can just use the slope!

You have to prove that there are/is atleast 1 point where they intersect!

You cannot simply plug in s = 3r +2. You have to determine whether the equations given have a root that is common with s = 3r + 2.

Statement 1:

s = 4r + 9 OR 3r + 2. Therefore insufficient.

Statement 2:

s = 4r - 6 OR 3r + 2. Also insufficient.

Statements 1 and 2: Taking both equations, we know that s MUST equal 3r + 2, since it is a root to both equations. Therefore C.

If you simply substitute s = 3r + 2, of course you'll get both sides equal to zero, because it is one of the two roots of each equation. This does not necessarily mean that both statements are insufficient by themselves.

You cannot simply plug in s = 3r +2. You have to determine whether the equations given have a root that is common with s = 3r + 2.

Well it depends. You can plug in s = 3r +2. In essense what you are doing is illustrating that at that point the same values of s,r makes sense for both the equation. Though, you cannot say that either equation 1 or 2 is equivalent to 0.

However, if it is equivalent to 0, you can show that the two equations are valid for the same values of s,r! I.e. the two curve/lines intersect.

Another way to explain the above would be to draw the graph of the two equations!

If they intersect, they will have the same values of s,r satisfying the two equations!

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

1, (3r+2-s)(4r + 9 -s) = 0

2, (4r -6 -s )(3r + 2 - s) = 0

A) (3r+2-s)(4r + 9 -s) = 0
=> (3r+2-s) = 0 or (4r + 9 -s)=0
so any pair of (r,s) which satisfy one of above will solve Eq-A
In sufficient

B) Also in sufficient with same logic

A+B to gather means only 3r+2-s=0 (s=3r+2) is valid option Hence C is the answer.

-------------------------------------------------
Other way to visualize this problem is to consider r and s as variables, this means 3r+2-s=0 is a Line(line1), so does 4r+9-s=0(line2), so any value on line 1 or line 2 will satifsy eq. one so we really dont know which exact line we are talking about.

Same logic will apply to B

But if we see A, B simultaniously only points on Line 3r+2-s=0 will satisfy both equation, hence C is the answer.

The official answer is wrong and I say this full authority.

In order for a point with coordinates (r,s) to lie on a line y=3x+2, the slope between the point (r,s) and another point lying on line y-3x+2 should be 3.

Lets say x =1 then y= 5

So slope between (R,S) and (1,5) should be 3.

Now lets take B

(4r -6 -s )(3r + 2 - s) = 0

Which means that 4r-s-s =0 and also 3r+2-s=0

Simply those so that

r= (s+6)/4 â€¦. (1)
s= s = 3r+2â€¦. (2)

now take the slope between (R,S) and (1,5)

slope = s-5 / r-1

substiture r and s from 1 and 2 and you will get:

slope = 12(r-1) / (s+2)

substitute s from 2

slope = 12(r-1) / 3r+4

you are stuck and you cannot find the slope. So B can not be the answer.

I´ve done an interview at Accepted.com quite a while ago and if any of you are interested, here is the link . I´m through my preparation of my second...

It’s here. Internship season. The key is on searching and applying for the jobs that you feel confident working on, not doing something out of pressure. Rotman has...