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Ch. 1 - Equations and Inequalities
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
Chapter 2, Problem 62

In Exercises 59–94, solve each absolute value inequality. |x + 3| ≤ 4

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1
Recall that the absolute value inequality \(|A| \leq B\) means that the expression inside the absolute value, \(A\), lies between \(-B\) and \(B\). So, rewrite the inequality \(|x + 3| \leq 4\) as a compound inequality: \(-4 \leq x + 3 \leq 4\).
Next, solve the compound inequality by isolating \(x\). Subtract 3 from all three parts of the inequality: \(-4 - 3 \leq x + 3 - 3 \leq 4 - 3\), which simplifies to \(-7 \leq x \leq 1\).
Interpret the solution: \(x\) is any number between \(-7\) and \(1\), inclusive, because the inequality is 'less than or equal to'.
Express the solution in interval notation as \([-7, 1]\), which represents all \(x\) values from \(-7\) to \(1\) including the endpoints.
Optionally, you can graph the solution on a number line by shading the region between \(-7\) and \(1\) and including solid dots at these points to indicate they are part of the solution.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Absolute Value Definition

The absolute value of a number represents its distance from zero on the number line, always as a non-negative value. For any real number x, |x| equals x if x is positive or zero, and -x if x is negative. Understanding this helps in interpreting and solving absolute value inequalities.
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Solving Absolute Value Inequalities

An inequality involving absolute value, such as |A| ≤ B, can be rewritten as a compound inequality: -B ≤ A ≤ B. This approach transforms the absolute value inequality into two linear inequalities that can be solved simultaneously to find the solution set.
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Compound Inequalities

Compound inequalities involve two inequalities joined by 'and' or 'or'. For absolute value inequalities like |x + 3| ≤ 4, the solution requires finding all x values that satisfy both -4 ≤ x + 3 and x + 3 ≤ 4 simultaneously, resulting in a range of solutions.
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