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Solve each inequality ||2x-1|-2|>3

User Dodd
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1 Answer

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Answer:

Solution: x < -2 or x > 3

Interval notation: (-∞, -2) ∪ (3, ∞)

Explanation:

Given inequality:


||2x-1|-2| > 3

Apply the absolute rule:


\textsf > a$, $a > 0$ \;then \;$u > a$ \;or \;$u < -a$.

Therefore:


\textsf{Case 1}: \quad |2x-1|-2 > 3


\textsf{Case 2}: \quad |2x-1|-2 < -3

Solve each case independently.

Case 1

Isolate the absolute value on one side of the equation:


\begin{aligned}\implies |2x-1|-2&amp; > 3\\\implies |2x-1| &amp; > 5 \end{aligned}

Apply the absolute rule:


\textsfu.


\begin{aligned}\underline{\textsf{Equation 1}} &amp; &amp; \quad \quad\underline{\textsf{Equation 2}}\\2x-1 &amp; > 5 &amp; 2x-1 &amp; < -5\\2x&amp; > 6 &amp; 2x&amp; < -4\\x&amp; > 3 &amp; x&amp; < -2\end{aligned}

Therefore, x < -2 or x > 3.

Case 2

Isolate the absolute value on one side of the equation:


\begin{aligned}\implies |2x-1|-2&amp; < -3\\\implies |2x-1| &amp; < -1 \end{aligned}

As an absolute value cannot be less than zero, there is no solution for x∈R.

Solution

Solution: x < -2 or x > 3

Interval notation: (-∞, -2) ∪ (3, ∞)

User Meinhard
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