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Re: In the xy-plane, does the line with equation y = 3x + 2 contains the [#permalink]
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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
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Re: In the xy-plane, does the line with equation y = 3x + 2 contains the [#permalink]
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Bunuel - you make every question looks so simple. always awesome explanation.
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awesome explanation Bunuel and Soumanag......thanks....
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Re: In the xy-plane, does the line with equation y = 3x + 2 contains the [#permalink]
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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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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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Bunuel
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.
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Re: In the xy-plane, does the line with equation y = 3x + 2 contains the [#permalink]
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Bunuel
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.

Hope it's clear.
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Re: In the xy-plane, does the line with equation y = 3x + 2 contains the [#permalink]
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.
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Re: In the xy-plane, does the line with equation y = 3x + 2 contains the [#permalink]
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danzig
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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Re: In the xy-plane, does the line with equation y = 3x + 2 contains the [#permalink]
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If the point (r,s ) lies in the equation y= 3x+2 then after substituting r,s in the line equation, we get.
S= 3r + 2
Or 3r+ 2 – S =0

So our Question here is to find whether 3r+2-S = 0 or not? Yes or No
Statement1.
(3r+2−s)(4r+9−s)=0
From Statement 1, We can infer that either 3r+2-s =0 or 4r+9-s =0 or both could be zero. Hence, we cannot answer the Question with a definite Yes or No.
So Statement 1 is insufficient. Answer option A and D can be eliminated. Possible answer options are B, C and E.

Statement 2.
(4r−6−s)(3r+2−s)=0
From Statement 2, We can infer that either 4r−6−s =0 or 3r+2−s or both could be zero. Hence, we cannot answer the Question with a definite Yes or No
So, Statement 2 is also insufficient. Answer option B can be eliminated. Possible answer options are C and E.


Let’s Try combining both statements.
4r+9-s =0 => 4r-s =-9
4r−6−s =0 => 4r-s = 6
So, both 4r+9-s =0 and 4r−6−s =0 cannot be 0 at the same time. As 4r-s cannot be both -9 and 6.
Hence, we can conclude that in one of the statements, 3r+2−s will be definetly 0.
So, we can answer the question by combining the both statements.
Option C

Thanks,
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Hi IanStewart - i actually chose (E) on this one and not (C)

From (s1) :
s = 3r + 2
OR
s = 4r +9

From (s2) :
s = 3r + 2
OR
s = 4r -6

When you combine the statements, there are 3 options for "s"

(i) s = 3r + 2
or
(ii) s = 4r +9
or
(iii) s = 4r -6

I see : s = 3r + 2 is a common factor but why does that imply, we can just "drop" option (ii) and option (iii) as possible values of "s" ?

When we combine the statements, can we really say s DOES NOT EQUAL 4r +9 | S does not equal 4r -6 ?

I thought upon combining the statement, there are 3 options for the value of "S"

hence "E"
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jabhatta2
Hi IanStewart - i actually chose (E) on this one and not (C)

From (s1) :
s = 3r + 2
OR
s = 4r +9

From (s2) :
s = 3r + 2
OR
s = 4r -6

Everything is correct to this point. When we combine the statements, clearly one possibility is that s = 3r + 2. If s is not equal to 3r + 2, then from Statement 1, the only remaining possibility is that s = 4r + 9, and from Statement 2, the only remaining possibility is that s = 4r - 6. So if s is not equal to 3r + 2, s must be equal to both 4r + 9 and 4r - 6, but those are definitely two different numbers (4r + 9 is exactly 15 greater than 4r - 6), so s could never simultaneously be equal to both of them, which is why the only possibility is that s = 3r + 2, and the answer is C.

The answer could have been E here if there was a legitimate second possibility -- it's because 4r - 6 and 4r + 9 can never be equal that s = 3r + 2 needs to be true.
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gaurav2k101
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) \((4r-6-s)(3r+2-s)=0\)

(r,s) is a point on the XY plane e.g (1, 2) or (2, 8) or (3, 21) etc. There is a defined relation between r and s.
Say if (r, s) is (1, 2), we can say that r + s = 3, s - r = 1, 2r + s = 4 etc. Each of these expressions will take only one value because both r and s have defined values.
This is an important point to understand in this question.

We want to know whether (r, s) lies on y = 3x + 2 i.e. whether s = 3r + 2. So we want to know whether the value of 3r - s is -2.
Note that all points (r, s) that satisfy this condition (3r - s = -2) will lie on the line y = 3x + 2. So the line represents the locus of all such points. e.g. (1, 5), (2, 8), (3, 11) etc.

Coming back to the question, we want to figure out whether 3r - s is -2.

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

This tells us that either 3r - s = -2 or 4r-s = -9. At least one of these relations certainly holds.
Not sufficient alone.

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

This tells us that either 4r - s = 6 or 3r - s = -2. At least one of these relations certainly holds.
Not sufficient alone.

Using both statements, now we know that one of 3r - s = -2 and 4r-s = -9 certainly holds and one of 4r - s = 6 and 3r - s = -2 certainly holds.

Now there are two possibilities.
Either 3r - s = -2 in which case, all works out.
Or 3r - s is not equal to -2 and both 4r-s = -9 and 4r - s = 6 hold. But here is the problem: both 4r-s = -9 and 4r - s = 6 cannot hold. As discussed above, 4r-s can take only one value. It cannot be both -9 and 6. Hence this case is not possible.

Then 3r - s must be equal to -2 and hence (r, s) must lie on the line y = 3x + 2.

Answer (C)
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KarishmaB
When you combine, now there are two possibilities.
Either 3r - s = -2 in which case, all works out.
Or 3r - s is not equal to -2 and both 4r-s = -9 and 4r - s = 6 hold. But here is the problem: both 4r-s = -9 and 4r - s = 6 cannot hold. As discussed above, 4r-s can take only one value. It cannot be both -9 and 6. Hence this case is not possible.

Then 3r - s must be equal to -2 and hence (r, s) must lie on the line y = 3x + 2.

Answer (C)

Hi Karishma – not sure I agree with the red.

Per the red – you are implying – one of the two statements - S1 or S2 is lying

But we know the statement DON’T lie.

(S1) / (S2) will never give factually incorrect information.

(s1) or (S2) could give us -- INCOMPLETE or COMPLETE information, but never FACTUALLY INCORRECT information.
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jabhatta2
KarishmaB
When you combine, now there are two possibilities.
Either 3r - s = -2 in which case, all works out.
Or 3r - s is not equal to -2 and both 4r-s = -9 and 4r - s = 6 hold. But here is the problem: both 4r-s = -9 and 4r - s = 6 cannot hold. As discussed above, 4r-s can take only one value. It cannot be both -9 and 6. Hence this case is not possible.

Then 3r - s must be equal to -2 and hence (r, s) must lie on the line y = 3x + 2.

Answer (C)

Hi Karishma – not sure I agree with the red.

Per the red – you are implying – one of the two statements - S1 or S2 is lying

But we know the statement DON’T lie.

(S1) / (S2) will never give factually incorrect information.

(s1) or (S2) could give us -- INCOMPLETE or COMPLETE information, but never FACTUALLY INCORRECT information.

Note that S1 says that EITHER 3r - s = -2 OR 4r-s = -9 (at least one of them must be true)
If we decide that 4r-s cannot be -9, then we get that 3r - s must be -2. Hence S1 is satisfied. One of them is true.
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Re: In the xy-plane, does the line with equation y = 3x + 2 contains the [#permalink]
Bunuel , I hope you are well. I have a doubt in this, its asking if s=3r+2.
in st1 - we given equation, in this if we substitute this value of s = 3r+2 then we will get LHS=RHS hence satisfied and similarly in st2. Isn't both sufficient alone?
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KarMishh
Bunuel , I hope you are well. I have a doubt in this, its asking if s=3r+2.
in st1 - we given equation, in this if we substitute this value of s = 3r+2 then we will get LHS=RHS hence satisfied and similarly in st2. Isn't both sufficient alone?

It seems that you did not read the discussion above.

From the first statement, the equation (3r + 2 - s)(4r + 9 - s) = 0 does not necessarily imply that 3r + 2 - s = 0, which would lead to the conclusion that 3r + 2 = s. It could be the case that 4r + 9 - s = 0, and in this case, 3r + 2 - s ≠ 0. This reasoning also applies to the second statement.

Please review the solutions provided above and study them carefully for a better understanding.
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