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

what is the shortest approach to solving this question?



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Inequality 05 Jun 2010, 14:45
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gmatbull
|x-1| + |x+2| <= 5

what is the shortest approach to solving this question?
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\(|x-1|+|x+2|\leq{5}\)

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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Inequality 05 Jun 2010, 17:02
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gmatbull
|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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Am grateful, it's very clear now. +1 Kudos to you.
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Inequality 29 Jan 2012, 08:15
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\(|x-1|+|x+2|\leq{5}\)

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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Inequality 29 Jan 2012, 22:18
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gmatbull
|x-1| + |x+2| <= 5

what is the shortest approach to solving this question?
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Check these posts for a 20 sec non-algebra approach.

https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... edore-did/

https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... 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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