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Inequality

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Inequality [#permalink] New post 05 Jun 2010, 13:19
|x-1| + |x+2| <= 5

what is the shortest approach to solving this question?
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Re: Inequality [#permalink] New post 05 Jun 2010, 13:45
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gmatbull wrote:
|x-1| + |x+2| <= 5

what is the shortest approach to solving this question?


|x-1|+|x+2|\leq{5}

Two key points: x=-2 and x=1 (key points are the values of x when absolute values equal to zero), thus three ranges to check:
---------{-2}--------{1}---------

1. x<-2 (blue range) --> |x-1|+|x+2|\leq{5} becomes: -x+1-x-2\leq{5} --> -3\leq{x} --> -3\leq{x}<-2;

2. -2\leq{x}\leq{1} (green range) --> |x-1|+|x+2|\leq{5} becomes: -x+1+x+2\leq{5} --> 3\leq{5}, which is true. This means that inequality |x-1|+|x+2|\leq{5} is ALWAYS true when x is in the range -2\leq{x}\leq{1};

3. x>1 (red range)--> |x-1|+|x+2|\leq{5} becomes: x-1+x+2\leq{5} --> x\leq{2} --> 1<x\leq{2}.

The ranges above give the following final range for which given inequality is true: -3\leq{x}\leq{2}.

Hope it's clear.
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Re: Inequality [#permalink] New post 05 Jun 2010, 16:02
Bunuel wrote:
gmatbull wrote:
|x-1| + |x+2| <= 5

what is the shortest approach to solving this question?


|x-1|+|x+2|\leq{5}

Two key points: x=-2 and x=1 (key points are the values of x when absolute values equal to zero), thus three ranges to check:
---------{-2}--------{1}---------

1. x<-2 (blue range) --> |x-1|+|x+2|\leq{5} becomes: -x+1-x-2\leq{5} --> -3\leq{x} --> -3\leq{x}<-2;

2. -2\leq{x}\leq{1} (green range) --> |x-1|+|x+2|\leq{5} becomes: -x+1+x+2\leq{5} --> 3\leq{5}, which is true. This means that inequality |x-1|+|x+2|\leq{5} is ALWAYS true when x is in the range -2\leq{x}\leq{1};

3. x>1 (red range)--> |x-1|+|x+2|\leq{5} becomes: x-1+x+2\leq{5} --> x\leq{2} --> 1<x\leq{2}.

The ranges above give the following final range for which given inequality is true: -3\leq{x}\leq{2}.

Hope it's clear.

Am grateful, it's very clear now. +1 Kudos to you.
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Re: Inequality [#permalink] New post 29 Jan 2012, 07:15
|x-1|+|x+2|\leq{5}

there are four ways to open the absolute 1st and 2nd expression: ++, --, +-, -+
however, +- & -+ will lead to cancellation of 'x' so we are left with same signs.

++ gives us: x-1+x+2\leq{5} or x\leq{2}

-- gives us: -x+1-x-2\leq{5} or -3\leq{x}

thus, -3\leq{x}\leq{2}
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Re: Inequality [#permalink] New post 29 Jan 2012, 21:18
Expert's post
gmatbull wrote:
|x-1| + |x+2| <= 5

what is the shortest approach to solving this question?



Check these posts for a 20 sec non-algebra approach.

http://www.veritasprep.com/blog/2011/01 ... edore-did/

http://www.veritasprep.com/blog/2011/01 ... s-part-ii/

(You need to go through the two posts to make sense of what I have written here.)

We want that the sum of 'distance of x from 1' and 'distance of x from -2' should be less than or equal to 5.
1 and -2 are 3 units away from each other. To get the exact sum of 5, we need to move 1 unit to the left or right. If we move one unit to the left, we get -3. One unit to the right takes us to 2. All points between them are acceptable since the sum is less than 5 in that region.
So you get -3 <= x <= 2
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Re: Inequality   [#permalink] 29 Jan 2012, 21:18
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