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HongHu
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MA
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prep_gmat
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MA
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from statement (I),
2r-s=-3 ............................. eq (1)
4r+2s=6 ........................... eq (2)
solving these equations yields r = 0 and s = 3
y=2x+3 => s=2r+3 => s=2(0)+3=3
satisfies the eq.



from statement (II),
3r+2s=5............................. eq (3)
2r-s=-3 ........................... eq (4)
solving these equations yields r = -1/7, s = 19/7
y=2x+3 => s=2r+3 => s=2(-1/7)+19/7=19/7
also satisfies the eq.......

OA is D....
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banerjeea_98
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MA
from statement (I),
2r-s=-3 ............................. eq (1)
4r+2s=6 ........................... eq (2)
solving these equations yields r = 0 and s = 3
y=2x+3 => s=2r+3 => s=2(0)+3=3
satisfies the eq.



from statement (II),
3r+2s=5............................. eq (3)
2r-s=-3 ........................... eq (4)
solving these equations yields r = -1/7, s = 19/7
y=2x+3 => s=2r+3 => s=2(-1/7)+19/7=19/7
also satisfies the eq.......

OA is D....


MA, u can't assume that both of the eqn in state 1 is 0 at the same time i.e. either one of them can be 0 to make the eqn valid, so no solution is possible. Same goes for eqn 2. U have to combine them to get an ans.
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MA
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banerjeea_98


MA, u can't assume that both of the eqn in state 1 is 0 at the same time i.e. either one of them can be 0 to make the eqn valid, so no solution is possible. Same goes for eqn 2. U have to combine them to get an ans.



baner,

Could you pls be more specific with some examples..........?
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toddmartin
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MA
banerjeea_98


MA, u can't assume that both of the eqn in state 1 is 0 at the same time i.e. either one of them can be 0 to make the eqn valid, so no solution is possible. Same goes for eqn 2. U have to combine them to get an ans.


baner,

Could you pls be more specific with some examples..........?


If you have 2 things multiplied together you don't know for sure they are both 0. For example, you could have the equation (x)*(0)=0

Now you know that one term or both terms are 0, but you don't know that x=0. x could = 50 or anything and the equation would still be true.
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So one cant multiply both equations together and then solve and check? I do understand that we dont know which one equals zero, so is that a good enough reason to answer E? Or are more calculations needed?
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HongHu
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OA is E, prep_gmat's explanation is correct.
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MA
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HongHu
OA is E, prep_gmat's explanation is correct.


Honghu,

explanation pls..........
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Quote:
Is point (r,s) on the line y=2x+3?

Meaning, is 2r-s+3=0?

(I) (2r-s+3)(4r+2s-6)=0
=>
2r-s+3=0 or
4r+2s-6=0 =>2r+s-3=0
insufficient
eg: (1,-1) is on the line while (1,1) isn't, both satisfy (I)

(II) (3r+2s-5)(2r-s+3)=0
3r+2s-5=0
or
2r-s+3=0
insufficient
eg: (1,-1) is on the line while (1, 1) isn't, both satisfy (II)

Combine (I) and (II), the same examples can still be used to satisfy both combined, while (1,-1) is on the line while (1,1) is not.
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I know I am late.... the answer will be E.
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thanks HongHu.. thats a good explanation..



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