/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 61 Solve each absolute value inequa... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solve each absolute value inequality. $$|x+3| \leq 4$$

Short Answer

Expert verified
The solution set is \(x \in [-1, 1]\).

Step by step solution

01

Resolve Absolute Value

The original inequality can be rewritten as two separate inequalities: \(x + 3 \leq 4\) and \(-(x + 3) \leq 4\). These inequalities cover the two possibilities that the absolute value allows for, and both must be solved independently.
02

Solve First Inequality

Subtract 3 from both sides of the first inequality \(x+3 \leq 4\) to isolate \(x\). This leads to the solution \(x \leq 1\).
03

Solve Second Inequality

Subtract 3 from both sides of the second inequality \(-(x+3) \leq 4\). This leads to \(-x \leq 1\). Multiply both sides by \(-1\) to solve for \(x\). When multiplying or dividing an inequality by a negative number, the direction of the inequality changes, leading to \(x \geq -1\).
04

Combine Results

The results of both inequalities are combined to form the final solution. Therefore, the solution to the inequality \(|x+3| \leq 4\) is \(x \in [-1, 1]\) because it satisfies both \(x \leq 1\) and \(x \geq -1\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.