Faster way to solve?

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Faster way to solve?

by aznmexicana » Tue Jul 20, 2010 10:26 pm
Where would one start in solving this?

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

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


Thanks in advance :)
Source: — Data Sufficiency |

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by kvcpk » Tue Jul 20, 2010 11:32 pm
1) (3r + 2 - s)(4r + 9 - s) = 0
Either of (3r + 2 - s) or (4r + 9 - s) should be 0
If 3r+2-s=0 answer is YES
If 4r + 9 - s is 0 then answer is NO
INSUFF

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

Combining:
4r + 9 - s and 4r - 6 - s
cant both be 0 at the same time because
4r +9 - s = 4r - 6 - s +15
Hence 3r+2-s should be 0.

pick C

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by Patrick_GMATFix » Wed Jul 21, 2010 6:48 am
aznmexicana wrote:In the xy-plane, does the line with the equation y=3x+2 contain point (r,s)?

1. (3r+2-s)(4r+9-s)=0
2. (4r-6-s)(3r+2-s)=0
The equation of the line is y=3x+2, where x and y represent the coordinate of any point (x,y) on that line. For point (r,s) to be on the line, it must be the case that s=3r+2.

Rephrase the question as: "Is s=3r+2?"

Statement (1) tells you that
---> 3r+2-s=0 --> s=3r+2 OR
---> 4r+9-s=0 --> s=4r+9.

Because we don't know whether s is 3r+2 or 4r+9, we cannot answer the rephrase. Statement (2) works the same way. You must merge the statements to solve because once the statements are merged the only value of s that agrees with both statements is s=3r+2. C is the answer.

You can view a much more detailed explanation and a step-by-step video solution at GMATPrep Question 1088. To practice similar questions in timed drills, set topic='Geometry, Coordinate' and difficulty='600-700' in the Drill Generator

Hope that helped,
-Patrick

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