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September 1st, 2016, 07:56 AM   #1
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Triangle Inequality: Prove Absolute Value Inequality

Help Please!


The Triangle Inequality Theorem states that ||x|-|y||<|x+y|<|x|+|y|

Given that |x-2|<(1/3)
Prove 4 < |3x-11| < 6

The solution for x is 5/3 < x < 7/3. But I cannot just say "as x is approaching" these values since using the proof is required. Anyone has an idea on how to manipulate |3x-11| to fit into the theorem?
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September 1st, 2016, 06:58 PM   #2
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Quote:
Originally Posted by StillAlive View Post
Help Please!


The Triangle Inequality Theorem states that ||x|-|y||<|x+y|<|x|+|y|

Given that |x-2|<(1/3)
Prove 4 < |3x-11| < 6

The solution for x is 5/3 < x < 7/3. But I cannot just say "as x is approaching" these values since using the proof is required. Anyone has an idea on how to manipulate |3x-11| to fit into the theorem?
Start with $\displaystyle \dfrac{5}{3} < x < \dfrac{7}{3}$. Multiply by 3 and subtract 11.
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September 1st, 2016, 09:09 PM   #3
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Alternatively, start with $|x-2|< \frac13 \implies -\frac13 < x-2 < \frac13$.

Now use normal algebraic manipulations to transform the $x-2$ term into $3x - 11$.
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September 2nd, 2016, 02:08 PM   #4
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you can also start by replacing the second 4<|3x-11|<6 by 4<3x-11<6 or -6<3x-11<-4 which gives
5<x<17/3 or 5/3<x<7/3.

now let's return to the first
|x-2|<1/3 gives 5/3<x<7/3.

conclusion
if 5/3<x<7/3 then 5<x<17/3 or 5/3<x<7/3.
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September 2nd, 2016, 05:42 PM   #5
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The Triangle Inequality Theorem states that ||x|-|y||<|x+y|<|x|+|y|
Shouldn't each "<" be "$\small\leqslant$"?
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September 2nd, 2016, 11:45 PM   #6
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of course. if x=y=0 we have equalities.
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