How do you solve $∣ ∣ ∣ \frac{1}{3} x + 2 ∣ ∣ ∣ - 8 > - 3$ $|1/3x + 2| - 8 > -3$ ?

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How do you solve $∣ ∣ ∣ \frac{1}{3} x + 2 ∣ ∣ ∣ - 8 > - 3$ $|1/3x + 2| - 8 > -3$ ?

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Answer 1 · 2016-07-20T04:25:49

x > 9
x < - 21

Explanation:

$I \frac{x}{3} + 2 I > - 3 + 8$ $Ix/3 + 2I > -3 + 8$
$I \frac{x}{3} + 2 I > 5$ $Ix/3 + 2I > 5$
separate solving into 2 parts:
a. $(\frac{x}{3} + 2) > 5$ $(x/3 + 2) > 5$
$\frac{x}{3} > 3$ $x/3 > 3$
$x > 9$ $x > 9$
b. -(x/3 + 2) > 5
-x/3 - 2 > 5
-x/3 > 7
x < - 21
Graph

============ -21------------------- 0---------- 9==============

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How do you solve $∣ ∣ ∣ \frac{1}{3} x + 2 ∣ ∣ ∣ - 8 > - 3$ $|1/3x + 2| - 8 > -3$ ?

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