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
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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

Bunuel - you make every question looks so simple. always awesome explanation.
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i was doing the same mistake of solving for r and s and then putting it in the original equation... thanks for the clarification.....

Thanks a ton Bunuel!

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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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.

If the point lies on the line, then it will satisfy the line equation.
that is -> s = 3r + 2
or to say 3r + 2 - s = 0

From 1 we get.
(3r + 2 - s) = 0 or (4r + 9 - s) = 0. NS
As (r,s) can lie on both lines, or on any one of then.

From 2 we get.
(3r + 2 - s) = 0 or (4r - 6 - s) = 0. NS
As (r,s) can lie on both lines, or on any one of then.

Combining both, we have only one root common.
(3r + 2 - s) = 0.

Ans C

Try to search using key words of the question. For example I'd search for: "equation", "contain", and "point". Though it's not a big problem if you search and don't find a question, moderators will help you with that. So, don't worry about that.
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PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!

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