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inequalities

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by resilient » Thu Apr 03, 2008 7:54 pm
How many possible integer values are there for x, if Absolute value 4x-3 <6?

a.1
b.2
c3
d4
e5


fairly simple question when set up this way:

-6 <4x-3<6 this will drive to the correct answer..


however, my initial approach was wrong and I would like to know why?

pos: 4x-3<6 = x< 9/4

neg: 4x-3<-6 = x<-3/4

thus, both answer give infinite negative digits? what went wrong?
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Re: inequalities

by Stuart@KaplanGMAT » Thu Apr 03, 2008 8:47 pm
Enginpasa1 wrote:How many possible integer values are there for x, if Absolute value 4x-3 <6?

a.1
b.2
c3
d4
e5


fairly simple question when set up this way:

-6 <4x-3<6 this will drive to the correct answer..


however, my initial approach was wrong and I would like to know why?

pos: 4x-3<6 = x< 9/4

neg: 4x-3<-6 = x<-3/4

thus, both answer give infinite negative digits? what went wrong?
The problem is that you divided by a negative and didn't realize it.

The two solutions are actually:

+(4x-3) < 6

and

-(4x-3) < 6

and when we "move" the negative sign over in the second inequality, we're really multiplying both sides by -1, which reverses the inequality.

So, the second inequality becomes:

4x -3 > -6
4x > -3
x > -(3/4)
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I get it

by resilient » Thu Apr 03, 2008 10:19 pm
Crystal Clear. Now I understand! WHat method should I employ in order to never repeat the problem again?
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Re: I get it

by Stuart@KaplanGMAT » Fri Apr 04, 2008 9:44 am
Enginpasa1 wrote:Crystal Clear. Now I understand! WHat method should I employ in order to never repeat the problem again?
Whenever you see absolute value and inequalities, be very careful. Also, whenever you "cancel out" or "move over" a number, be aware that what you're really doing is dividing or multiplying both sides of the equation (or inequality) by that number. Especially when inequalities are involved, doing so can have unexpected consequences.
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by sampleresume » Sun Apr 06, 2008 5:15 pm
The answer is five. X can be any value of -2, -1, 0, 1, and 2.
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by senthil » Mon Apr 07, 2008 1:52 am
could you please explain me why have we taken this inequality...
-6 <4x-3<6 .....


Thanks
Senthil
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by resilient » Mon Apr 07, 2008 8:36 am
Senthil:

we start with this in absolute form: 4x-3 <6 SO whenever we see absolutes we need to worry about the pos and neg outcome.

pos:
4x-3<6

neg: -4x+3>6 (> since we are charging the whole situation by negative.) This changes the sign. After that, we come to your final solution. I hope that helps
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by GMAT_crusher » Mon Apr 07, 2008 7:08 pm
Still missing something here
checking for -6 < 4x-3 < 6

x for (4x - 3)
1 = 1
2 = 5
0 = -3

3 = 9
-1 = -7
-2 = -11
-3 = -16

total x values fit is 3!! what am I missing!!!
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ok

by resilient » Mon Apr 07, 2008 7:36 pm
the golden rule to understand is to know that with absolutes you need to account for a positive outcome and negative outcome. On top of this you need to make sure that whenver you divide by a negative you have to change the sign of the inequality! I suggest you read up a bit on inequalities and absolutes. Try to fullyunderstand and synthesize Stuarts explanation!
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by mmukher » Mon Apr 07, 2008 7:47 pm
Still missing something here
checking for -6 < 4x-3 < 6

x for (4x - 3)
1 = 1
2 = 5
0 = -3

3 = 9
-1 = -7
-2 = -11
-3 = -16

total x values fit is 3!! what am I missing!!!

you were almost there. The question states that x is an integer, so you need to ignore 0.

The actual answer was never stated clearly. Its B.

-3/4 < x < 2.5 so x can be 1,2 (not 0)
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by sandeepdeb » Tue Apr 08, 2008 1:11 am
Can anyone please tell the corretc answer? I though 0 is an integer and the answer to this question is 3. X lies between -3/4 and 9/4 and so, the possible integer values are 0,1 and 2.
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by mmukher » Tue Apr 08, 2008 7:53 am
My Bad. 0 is an integer. Possible values are 0,1,2 Answer is C.3
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