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Given a<0, what is another way of expressing -ax<1?

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Senior Manager
Joined: 15 Feb 2018
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Given a<0, what is another way of expressing -ax<1?  [#permalink]

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16 Jan 2019, 06:48
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Difficulty:

55% (hard)

Question Stats:

47% (01:05) correct 53% (01:02) wrong based on 41 sessions

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Given a<0, what is another way of expressing -ax<1?

A) x<1/a
B) x>1/a
C) x<-1/a
D) x>-1/a
E) x<1
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Joined: 02 Sep 2009
Posts: 53657
Re: Given a<0, what is another way of expressing -ax<1?  [#permalink]

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16 Jan 2019, 06:54
philipssonicare wrote:
Given a<0, what is another way of expressing -ax<1?

A) x<1/a
B) x>1/a
C) x<-1/a
D) x>-1/a
E) x<1

Since a is negative, then -a is positive, so we can divide the given inequality and keep the sign: x < -1/a.

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Re: Given a<0, what is another way of expressing -ax<1?  [#permalink]

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16 Jan 2019, 06:56
Bunuel, how is it different to b?
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Joined: 13 Feb 2018
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GMAT 1: 640 Q48 V28
Re: Given a<0, what is another way of expressing -ax<1?  [#permalink]

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16 Jan 2019, 07:03
1
philipssonicare wrote:
Bunuel, how is it different to b?

You multiply only right hand side by -1 change the sign and leave left hand side as is

x<-1/a multiply by -1 is -x>1/a
not
x>1/a
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Posts: 53657
Re: Given a<0, what is another way of expressing -ax<1?  [#permalink]

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16 Jan 2019, 07:04
1
philipssonicare wrote:
Bunuel, how is it different to b?

B reads: $$x > \frac{1}{a}$$.

The correct answer is $$x < -\frac{1}{a}$$, which can also be written as $$-x > \frac{1}{a}$$ (by multiplying it by -1). As you can see B and C are not identical.
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Re: Given a<0, what is another way of expressing -ax<1?   [#permalink] 16 Jan 2019, 07:04
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