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How about this one

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by Brent@GMATPrepNow » Tue Dec 16, 2008 5:51 pm
What is the value of x?

(1) 3(x–1) = y+3x
(2) (x+2)(y+3) = 0

Answer: E
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by cramya » Tue Dec 16, 2008 5:55 pm
Stmt I

3(x-1) = y+3x

3x-3 = y+3x

y=-3
No way to find X

INSUFF


Stmt II

(x+2)(y+3) = 0

Either x ix-2 or y is -3 or they both can be -2 and -3

INSUFF

Stmt I and II together

y=-3

Still we dont know if x is -2 or any other value

INSUFF

Choose E)
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Re: How about this one

by logitech » Tue Dec 16, 2008 7:25 pm
(1) 3(x–1) = y+3x

3x cancels. no info for X INSUF

(2) (x+2)(y+3) = 0

X=-2 or y=-3 INSUF

(1)+(2)

Can not be solved for X because there is no X values in statement 1

Choose E
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by vittalgmat » Wed Dec 17, 2008 12:02 am
in stmt 1 we cant find value of x.
Is it coz x can potentially have infinite values??

thanks
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by parallel_chase » Wed Dec 17, 2008 2:47 am
vittalgmat wrote:in stmt 1 we cant find value of x.
Is it coz x can potentially have infinite values??

thanks
Yes thats correct

3(x–1) = y+3x

3x - 3 = y + 3x

y = -3

substitute value of y back in the equation

3x - 3 = -3 + 3x

LHS = RHS

Therefore, any value of x will satisfy the equation. Hence, we cannot get a single value of x.
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Re: How about this one

by 4meonly » Wed Dec 17, 2008 3:00 am
Brent Hanneson wrote:What is the value of x?

(1) 3(x–1) = y+3x
(2) (x+2)(y+3) = 0

Answer: E
Brent, GIVE US MORE NEW TOUGH QUESTIONS!!!! :D :D :D
Your questions are very interesting. I solved all questions from your site :D
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Re: How about this one

by vittalgmat » Wed Dec 17, 2008 10:06 am
4meonly wrote:
Brent Hanneson wrote:What is the value of x?

(1) 3(x–1) = y+3x
(2) (x+2)(y+3) = 0

Answer: E
Brent, GIVE US MORE NEW TOUGH QUESTIONS!!!! :D :D :D
Your questions are very interesting. I solved all questions from your site :D
I second that!!!

Yeah.. Brent pllllllllleeeeeeaaaaase give us new tough questions.. I love your your approach to solutions as well...
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by ronniecoleman » Wed Dec 17, 2008 7:44 pm
Thanks Brent!! For nice questions!!
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by schumi_gmat » Thu Dec 18, 2008 1:24 pm
This a good question.

I have one question regarding stmt B.
from B, we know x=-2 and y=-3

Are we discarding B, because there are 2 solns for the equations regardless of we have unique values of x and y?
Cramya wrote:

Stmt II

(x+2)(y+3) = 0

Either x ix-2 or y is -3 or they both can be -2 and -3
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by Brent@GMATPrepNow » Thu Dec 18, 2008 1:35 pm
I have one question regarding stmt B.
from B, we know x=-2 and y=-3

Are we discarding B, because there are 2 solns for the equations regardless of we have unique values of x and y?

Quote:
Cramya wrote:

Stmt II

(x+2)(y+3) = 0

Either x ix-2 or y is -3 or they both can be -2 and -3
Actually, there are an infinite number of solutions to the equation (x+2)(y+3) = 0
Here are a few:
y = -3 and x=7
y = -3 and x=-4
y = -3 and x=222
y = -3 and x=1,000,000
etc.
If you plug in these values, it is true that (x+2)(y+3) = 0
Since x can have an infinite number of values, we can't definitively say what x is.

I hope that helps.
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by iamcste » Thu Dec 18, 2008 1:43 pm
schumi_gmat wrote:This a good question.

I have one question regarding stmt B.
from B, we know x=-2 and y=-3

Are we discarding B, because there are 2 solns for the equations regardless of we have unique values of x and y?
Cramya wrote:

Stmt II

(x+2)(y+3) = 0

Either x ix-2 or y is -3 or they both can be -2 and -3

eqtn is satisifed if

1. x=-2, y=irrespective, or

2. y=-3, x=irrespective, or

3. x=-2 and y=-3

so, we dont know exact value of x ( from above, it may be -2 or any other value when y=-3)

Insufficient
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