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inequality 1

Expert replies
Source: — Data Sufficiency |

by DanaJ » Tue Feb 03, 2009 3:39 am
1. y - x = 3 is equivalent to y = x +3. Now replace this little thing over there and you get that:
6x(x + 3) = x^2(x + 3) + 9(x + 3) or
6x^2 + 18 x = x^3 + 3x^2 + 9x + 27
x^3 - 3x^2 - 9x + 27 = 0.
These are my favorite type of questions since you have to be creative and notice the connections:
x^3 - 3x^2 - 9x + 27 = (x^3 - 9x) - (3x^2 - 27) = x(x^2 - 9) - 3(x^2 - 9). As you can see, x^2 - 9 is a common factor, so we get that:
(x - 3)(x^2 - 9) = 0. Now we go on and we get that, since x^2 - 9 = (x - 3)(x + 3), (x-3)(x + 3)(x - 3) = 0. This equation has two roots, 3 and -3, so we cannot say for sure which is the correct answer.

2. Again, get creative!
6xy = x^2y + 9y is equivalent to x^2y - 3xy - 3xy + 9y = xy(x - 3) - 3y (x - 3) = y(x - 3)(x - 3) = 0. Now, there are only to possible ways that this thing could equal 0, and that's if:
a. x - 3 = 0 or x = 3. This is impossible since x^3 < 0 and 3^3 = 27
b. y = 0.

If y = 0 , then xy = 0.

Answer B
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by willbeatthegmat » Tue Feb 03, 2009 9:36 am
I think stat 1 is also sufficient to answer this question.here is how....

6x= y(x^2 + 9)/y

x^2+9-6x= 0 or (x-3)^2 = 0
therefore, x=3.. so we just need value of y to find out xy
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by sfd2102 » Tue Feb 03, 2009 11:35 am
The equation does not have two solutions. Only x=3 is correct. Not x=-3
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by peddisetty » Tue Feb 03, 2009 5:38 pm
Yes SFD is correct. The first one doesnt have two solutions. y cancels on both sides.

Dana: Please clarify. Thanks
Raj Peddisetty
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by Alara533 » Tue Feb 03, 2009 8:02 pm
Dana is correct.
'willbeatthegmat' canceled 'y' assuming that 'y' not equal to zero. What if y is zero?

So we have to take care of that situation as well.

Answer is B.
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