In the xy-plane, does the line with equation y = 3x + 2 contains the

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.
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I took it totally wrong:
What i did was , from stmt 1:
1) (3r+2-s)(4r+9-s)=0
3r+2-s=0
4r+9-s=0 and equating both of them to find value of r and s.

Total disaster.

Thanks a lot guys ...Kudos to Bunuel and soumanag

Substituting (r,s) in the original equation:
y=3x+2 => s=3r+2 => 3r-s+2=0.
Question : is 3r-s-2=0. ??
statement 1 - (3r+2-s)(4r+9-s)=0 => (3r+2-s)=0 or (4r+9-s)=0
Not sufficient

Statement 2 - (4r-6-s)(3r+2-s)=0 => (3r+2-s)=0 or (4r-6-s)=0
Not sufficient
Combining 1&2 ->
common result (3r+2-s)=0
Sufficient.
Answer - C

Bunuel - you make every question looks so simple. always awesome explanation.
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awesome explanation Bunuel and Soumanag......thanks....

y = 3x +2 will contain the point (r,s), if :
s = 3r +2 (substitute r for x and s for y , in the eqn of line) => if 3r+2 -s = 0 ?

Statement 1 :
(3r + 2 - s) (4r + 9 - s) = 0
=> either (3r + 2 -s) =0 or (4r +9 -s ) = 0 -> Not Sufficient

Statement 2 :
(4r -6 -s) (3r + 2 - s) = 0
=> either (4r -6 -s) = 0 or (3r + 2 -s ) = 0 -> Not sufficient

Combining them :

3r + 2 - s = 0 - Sufficient (As [4r -s +9] and [4r -s -6] both can't be zero at the same time, they have a difference of 15).

Hence , C.

s = 3r + 2 should be true if equation y = 3x+2 contain the point (r,s)

From (1), s = 3r + 2 could be possible, but not sure

because 4r + 9 - s = 0 or 4r - s = -9 might be possible as well.


From(2), s = 3r + 2 could be possible, but not sure

because 4r - 6 - s = 0 or 4r - s = 6 might be possible as well.

However, from (1) and (2), only 3r + 2 - s = 0 can be true

else -9 = 6, which is absurd. So the answer is C.
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Ok.. So lets look at both the statements individually:

First Statement 1:

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

Look at the statement. It could mean two things. Either (3r+2-s) = 0 or (4r+9-s) = 0

if (3r+2-s) = 0, then yes 3r+2-s = 0 and hence s=3r+2 which is exactly like saying y=3x+2

But if 3r+2-s is not equal to 0 then we don't know for sure.


Now lets look at statement 2:

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

Same implies for statement 2.

now lets assume that in the first statement 4r+9-s=0. If that is true, then from the second statement it is not possible for 4r-6-s to be equal to 0. Hence 3r+2-s = 0 and yes the point r,s which is the same as x,y lies on it !

Hence C

Easy one. GMAT loves these type of patterns.

NOTE: Notice the pattern here. When you have similar type of information in both statements, the answer is usually C, D or E.

1. rephrasing the equations gives us s=3r+2 and s=4r+9. obviously s=4r+9 does not lie, but s=3r+2 does
2. rephrasing the equations gives us s=3r+2 and s=4r-5. obviously s=4r-5 does not lie, but s=3r+2 does

Together, s=3r+2 and it does lie. C.

Bunuel,

Again, sorry for re-opening an old thread but I have a small doubt regarding this question. I got this question on my GMAT Prep today, had one quick look at the options and immediately saw that for the point to lie on the line, 3r+2-s has to be equal to zero. And since this information was given to us in two separate quadratic equations, it was necessary for us to have both the equations in order to discard (4r+9-s) and (4r-6-s) as solutions to the given equations.

However, at this point, I thought to myself that C would be too easy an answer and solved the equations to see if GMAT was upto its usual trickery. And this is where I went wrong.

Take for example the first equation. One solution gives us what we want, i.e., 3r+2-s = 0. However, if we solve the second solution 4r+9-s = 0, we get s = 4r+9. Now, if we plug this into the original equation y = 3x+2, we get:

y = 3x+2
Plugging the values of r and s into x and y:

4r+9 = 3r +2;
Solving this gives us r = -7 and s = -19 and plugging these values of r and s in the original equation satisfies the equation:

-19 = 3(-7) + 2;
-19 = -19

Since, this equation too satisfies the equation (and similarly, so does the second equation) I marked the answer as D.

Can you please tell me where I am going wrong?

Many thanks in advance.

The question asks whether \(3r+2-s=0\). Now, statement (1) says: \((3r+2-s)(4r+9-s)=0\), which means that either \(3r+2-s=0\) OR \(4r+9-s=0\) OR both.

The above DOES NOT mean that we have a system of equations: \(3r+2-s=0\) and \(4r+9-s=0\). Because if it were so then you have an answer right away: \(3r+2-s=0\). No wonder that when you solve it as a system you got the values of \(s\) and \(r\) which make \(3r+2-s\) equal to zero: your whole starting point was that it does.

Hope it's clear.
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A*B=0==>THEN EITHER A/B IS ZERO OR BOTH ARE ZERO...this is a rule.
now if r,s lies on point y=3x+2==>then 3r+2-s=0
now lets say 3r+2-s=A..
4r+9-s=B
4r-6-s=C

now statement 1 SAYS that ...A*B=0===>Clearly we are not sure that A=0 or not
similarly statement 2 says that ...A*C===>AGAIN WE CANNOT GAURANTEE THAT A=0..

NOW combining both we cam clearly say that A=0..
HENCE SUFFICIENT.

Bunuel,

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.

Hope it's clear.
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(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.

Yes, in this case the answer would be E.
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Required: Does the line with equation y=3*X +2 contain the point (r,s)
Or simply put, Is 3r - s +2 = 0?

Statement 1: (3r + 2 - s)(4r + 9 - s) = 0
This means either
(3r + 2 - s) = 0
Or (4r + 9 - s) = 0
INSUFFICIENT

Statement 2: (4r - 6 - s)(3r + 2 - s) = 0
This means either
(4r - 6 - s) = 0
Or (3r + 2 - s) = 0
INSUFFICIENT

Combining both statements:
We know that (3r + 2 - s) = 0
SUFFICIENT

Correct Option: C

(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)

so only case 1 holds i.e.

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


(C) is the answer.
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I thought this one was one of the harder questions on the official practice exam #1. However, I got it correct by using the number of variables approach, i.e. we need both the equations because we have two variables. Im not sure how valid that approach actually is on this question but I couldnt solve it using the concepts explained here.

How do we know that both parentheses cant be 0? I mean, what if (3r+2−s) = (4r+9−s) = 0 ? Is this completely impossible?

(1) (3r+2−s)(4r+9−s)=0

Bunuel: How do you know that either of the parentheses are 0? What if both are?