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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.

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

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04 Sep 2010, 07:19

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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

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

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10 Jan 2012, 08:17

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

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10 Jan 2012, 13:06

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

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20 Apr 2012, 06:25

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 \(4r-6-s=0\) can not 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.

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.

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 \(4r-6-s=0\) can not 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.

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.

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

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08 Jul 2013, 03:59

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gaurav2k101 wrote:

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

(1) (3r+2-s)(4r+9-s)=0 (2) (4r-6-s)(3r+2-s)=0

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

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30 Aug 2013, 15:41

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.

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 contains the [#permalink]

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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

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