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question about absolute values

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question about absolute values

by jamesk486 » Mon Aug 18, 2008 10:38 pm
Is |x|>1?
(1) |x|>3
(2)|x+8| > 3 |x-2|

(1) is sufficient
but i had a question about (2)
so there are 4 cases, (i) when |x+8| >0, (ii) |x+8|<0, (iii) |x-2|<0, and (iv) |x-2| >0

so then is this how you solve it?
case (i) |x+8| > 3 |x-2| (both are positive)
case (ii) -|x+8| > 3 |x-2|
case (iii) -|x+8| > -3 |x-2| (both are negative)
case (iv) |x+8| > -3 |x-2|

is this right??
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by gmatters2vj » Mon Aug 18, 2008 10:48 pm
Hi James,
Do u really need to solve it?

For case 2 i have the following explanation:

|x+8| > 3 |x-2|

Please note that this condition is true for x=0 i.e.
when x=0, 8>6. For this |x|<1.
But when we take x=2
we get 10>0 which is true. For this |x|>1.

Thus insufficient.

Answer must be A.
Whats the answer?

Thanks
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by maihuna » Mon Apr 13, 2009 2:50 am
Ian,
Canyou please comment on a proper way to solve part 2?

I came up with following concludion:

-------------|-----------------------------|-------------
-8 2

For x<-8, -(x+8) > -3(x-2) OR x>-1
-8<x<2, x+8 > -3(x-2) or x>-1/2
x>2 x+8 > 3(x-2) or x<7

question is how to come to an conclusion, is my line no correct
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by Ian Stewart » Mon Apr 13, 2009 6:34 am
maihuna wrote:Ian,
Canyou please comment on a proper way to solve part 2?

I came up with following concludion:

-------------|-----------------------------|-------------
-8 2

For x<-8, -(x+8) > -3(x-2) OR x>-1
-8<x<2, x+8 > -3(x-2) or x>-1/2
x>2 x+8 > 3(x-2) or x<7

question is how to come to an conclusion, is my line no correct
Yes, your analysis is correct, though I've highlighted one line above: when you assumed that x < -8, you then found x > -1. The conclusion contradicts your assumption (x can't both be less than -8 and greater than -1), so this case is impossible, and you should ignore it. The other two cases are perfectly consistent, and combining their conclusions, you have that -1/2 < x < 7.

If you use the fact that |a - b| just represents the distance between a and b on the number line, we can rewrite Statement 2 as:

|x - (-8)| > 3 |x - 2|

In words, this says "x is more than three times as far from -8 as it is from 2". So, x must be closer to 2 than to -8. That's another way to see that x cannot possibly be less than -8, so we don't need to consider that case.
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by aj5105 » Thu Apr 30, 2009 5:05 am
Ian,

Needed your help on the below logic.

Statement 2:
|x + 3| > 3 |x - 2|

3 is positive. |x - 2| has to be positive too. Hence |x + 3| has to be positive.

Now,

x - 3 > 3 (x-2)

Simplify to get x < 7.

Insufficient.

Am i right here?
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by Ian Stewart » Fri May 01, 2009 12:25 pm
aj5105 wrote:Ian,

Needed your help on the below logic.

Statement 2:
|x + 3| > 3 |x - 2|

3 is positive. |x - 2| has to be positive too. Hence |x + 3| has to be positive.

Now,

x - 3 > 3 (x-2)

Simplify to get x < 7.

Insufficient.

Am i right here?
Your Statement 2 seems to be different from the one in the question above, but in any case, I don't follow the logic you're using here. I'm guessing you're replacing |x-2| with x-2 (removing the absolute value) because, as you say, |x-2| is positive? If that's what you were doing, that's not correct. |x-2| will always be positive (unless x = 2, in which case it equals zero). If we want to get rid of the absolute value, what we want to know is whether what's inside the absolute value is positive or negative:

|y| = y is y is positive
|y| = -y if y is negative

It's because of this that algebraic solutions to absolute value problems often require different cases. Here, for example, |x-2| = x-2 *if* x-2 is positive (i.e. if x > 2). On the other hand, |x-2| = -(x-2) if x-2 is negative (i.e. if x < 2).
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by maihuna » Fri May 01, 2009 12:31 pm
Ian I am confused now, does it mean equal to zero is another case?
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by Uri » Fri May 01, 2009 1:23 pm

(1) is sufficient.
(2)|x+8| > 3 |x-2|
if (x-2)>0, then (x+8)>0
so, x+8>3(x-2)
simplifying we get, x<7
if (x-2)<0, then we can write,
x+8>3(2-x)
simplifying we get, x>-1/2
thus -1/2<x<7
so, |x| is not necessarily greater than 1. hence, insufficient.
Ans. (A)

i hate absolute value inequality problems. can anyone help us with some good tutorial on solving this type of problem?
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by Ian Stewart » Fri May 01, 2009 1:48 pm
maihuna wrote:Ian I am confused now, does it mean equal to zero is another case?
No, you can include zero in either case, since:

|y| = y if y > 0

|y| = -y if y < 0
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