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# Inequality

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Director
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05 Jun 2010, 14:19
|x-1| + |x+2| <= 5

what is the shortest approach to solving this question?
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05 Jun 2010, 14:45
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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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Director
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05 Jun 2010, 17:02
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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29 Jan 2012, 08:15
$$|x-1|+|x+2|\leq{5}$$

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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29 Jan 2012, 22:18
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, 22:18
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