To calculate: Thesolution set of the inequality 2&lt;−1+|4w−3| in the interval notation.

Given Information: The providedinequalityis 2&lt;−1+|4w−3|. Formula used: Fora real number k&gt;0, |u| &gt; k is equivalent tow &lt; —k or u&gt; k. Calculation: Consider the inequality, 2&lt;−1+|4w−3| Add | to both sides of the above equation. 2&lt;−1+4w−3 2+1&lt;−l+4w−3+1 3&lt;4w−3 Write the given equation in the form u &lt;−koru&gt;k 4w−3&lt;−3 4w&lt;−3+3 4w&lt;0 w&lt;0 or 4w−3&gt;3 4w&gt;3+3 w&gt;64 w&gt;32 Hence, the solution set of the inequality 2&lt;−1+|4w−3| in the interval notation is (−∞,0)∪(32,∞).

Question: "To calculate: Thesolution set of the inequality 2<−1+|4w−3|..."

High School
Other

Barbara Schroder

Answered question

2021-11-15

To calculate: Thesolution set of the inequality $2 < - 1 + | 4 w - 3 |$ in the interval notation.

Answer & Explanation

Blanche McClain

Beginner2021-11-16Added 15 answers

Given Information:
The providedinequalityis $2 < - 1 + | 4 w - 3 |$ .
Formula used:
Fora real number $k > 0$ ,
$| u |$ > k is equivalent tow < —k or u> k.
Calculation:
Consider the inequality,
$2 < - 1 + | 4 w - 3 |$
Add | to both sides of the above equation.
$2 < - 1 + 4 w - 3$
$2 + 1 < - l + 4 w - 3 + 1$
$3 < 4 w - 3$
Write the given equation in the form u $< - k or u > k$
$4 w - 3 < - 3$
$4 w < - 3 + 3$
$4 w < 0$
$w < 0$
or
$4 w - 3 > 3$
$4 w > 3 + 3$
$w > \frac{6}{4}$
$w > \frac{3}{2}$
Hence, the solution set of the inequality $2 < - 1 + | 4 w - 3 |$ in the interval notation is $(- \infty, 0) \cup (\frac{3}{2,} \infty)$ .

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