Solve absolute value inequality. 1&lt;|x-11/3|+7/3

Step 1 Given that, 1&lt;|x−113|+73 Then either ⇒1−73&lt;|x−113| ⇒3−73&lt;x−113 ⇒−43&lt;x−113 ⇒−43+113&lt;x Step 2 ⇒−4+113&lt;x ⇒73&lt;x Or −1&gt;|x−113|−73 ⇒−1+73&gt;x−113 ⇒−3+73&gt;x−113 Step 3 ⇒−43&gt;x−113 ⇒−43+113&gt;x ⇒−4+113&gt;x ⇒73&gt;x Result: So either x&lt;73 or x&gt;73

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elvishwitchxyp

Answered question

2021-12-11

Solve absolute value inequality.
1<|x-11/3|+7/3

Answer & Explanation

Barbara Meeker

Beginner2021-12-12Added 38 answers

Step 1
Given that,
$1 < | x - \frac{11}{3} | + \frac{7}{3}$
Then either
$\Rightarrow 1 - \frac{7}{3} < | x - \frac{11}{3} |$
$\Rightarrow \frac{3 - 7}{3} < x - \frac{11}{3}$
$\Rightarrow \frac{- 4}{3} < x - \frac{11}{3}$
$\Rightarrow \frac{- 4}{3} + \frac{11}{3} < x$
Step 2
$\Rightarrow \frac{- 4 + 11}{3} < x$
$\Rightarrow \frac{7}{3} < x$
Or
$- 1 > | x - \frac{11}{3} | - \frac{7}{3}$
$\Rightarrow - 1 + \frac{7}{3} > x - \frac{11}{3}$
$\Rightarrow \frac{- 3 + 7}{3} > x - \frac{11}{3}$
Step 3
$\Rightarrow \frac{- 4}{3} > x - \frac{11}{3}$
$\Rightarrow \frac{- 4}{3} + \frac{11}{3} > x$
$\Rightarrow \frac{- 4 + 11}{3} > x$
$\Rightarrow \frac{7}{3} > x$
Result: So either $x < \frac{7}{3} or x > \frac{7}{3}$

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