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

Question

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

Bunuel · Accepted Answer

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.

gaurav2k101 · Answer

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

soumanag · Answer

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

ichha148 · Answer

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

saxenashobhit · Answer

awesome explanation Bunuel and Soumanag......thanks....

omerrauf · Answer

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

shinbhu · Answer

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.

arjuntomar · Answer

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.

Bunuel · Answer

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.

danzig · Answer

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.

Bunuel · Answer

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.

CrackverbalGMAT · Answer

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,
Clifin J Francis,
GMAT SME

jabhatta2 · Answer

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"

IanStewart · Answer

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.

KarishmaB · Answer

(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)

jabhatta2 · Answer

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.

KarishmaB · Answer

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.

testtango · Answer

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?

Bunuel · Answer

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.

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

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