How do you solve and write the following in interval notation: $- \frac{1}{6} + ∣ ∣ 2 - \frac{x}{3} ∣ ∣ > \frac{1}{2}$ $-1/6+|2−x/3|>1/2$ ?

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How do you solve and write the following in interval notation: $- \frac{1}{6} + ∣ ∣ 2 - \frac{x}{3} ∣ ∣ > \frac{1}{2}$ $-1/6+|2−x/3|>1/2$ ?

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Answer 1 · 2018-04-19T14:09:23

$x \in [- \infty, 4) and x \in (8, + \infty]$ $x in[-oo,4)andx in(8,+oo]$ or $x \notin (4, 8)$ $x notin(4,8)$

Explanation:

First we rearrange to get the $| f (x) |$ $abs(f(x))$ part on its own by adding $\frac{1}{6}$ $1/6$ to both sides.

$∣ ∣ 2 - \frac{x}{3} ∣ ∣ > \frac{2}{3}$ $abs(2-x/3)>2/3$

Due to the nature of $| |$ $abs()$ we can take the inside to be positive or negative, since it turn either into a positive number.

$2 - \frac{x}{3} > \frac{2}{3}$ $2-x/3>2/3$ or $- 2 + \frac{x}{3} > \frac{2}{3}$ $-2+x/3>2/3$
$\frac{x}{3} < 2 - \frac{2}{3}$ $x/3<2-2/3$ or $\frac{x}{3} > \frac{2}{3} + 2$ $x/3>2/3+2$
$\frac{x}{3} < \frac{4}{3}$ $x/3<4/3$ or $\frac{x}{3} > \frac{8}{3}$ $x/3>8/3$
$x < 4$ $x<4$ or $x > 8$ $x>8$

So, we have $x \in [- \infty, 4) and x \in (8, + \infty]$ $x in[-oo,4)andx in(8,+oo]$ or $x \notin (4, 8)$ $x notin(4,8)$

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How do you solve and write the following in interval notation: $- \frac{1}{6} + ∣ ∣ 2 - \frac{x}{3} ∣ ∣ > \frac{1}{2}$ $-1/6+|2−x/3|>1/2$ ?

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Explanation:

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How do you solve and write the following in interval notation: $- \frac{1}{6} + ∣ ∣ 2 - \frac{x}{3} ∣ ∣ > \frac{1}{2}$ $-1/6+|2−x/3|>1/2$ ?