Law of Trichotomy Applied to Absolute Value Inequalities in One Variable Part IV

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Continuing information of the previous few posts.

Fact B: If k is a positive real number and M is a linear expression, then |M| < k is equivalent to the compound inequality – k < M < k.
Example: The inequality |2x – 3| < 7 is equivalent to –7 < 2x – 3 < 7.
Example: The inequality |5x – 8| < 3 is equivalent to -3 < 5x – 8 < 3.
Example: The inequality is equivalent to
Observe that in the general statement and each of the examples the expression inside the absolute value symbol is wedged between the number k and its opposite.

Fact C: If k is a positive real number and M is a linear expression, then the solution set for |M| < k is an interval on the real number line.
Example: The inequality |2x – 3| < 7 is equivalent to –7 < 2x – 3 < 7 which is equivalent to –4 < 2x < 10 which is equivalent to –2 < x < 5 whose solution is the interval (–2, 5).
It follows that:

Fact D: The endpoints of that interval are the solutions of the boundary equation |M| = k.
Example: (–2, 5) is the solution set for |2x – 3| < 7. Its endpoints are the solutions of the corresponding boundary equation |2x – 3| = 7. The solution set for the boundary equation |2x – 3| = 7 is {–2, 5}.
And finally it follows that:

Fact E: Every real number outside the interval and not an endpoint of the interval is a solution to |M| > k.
Example: (–2, 5) is the solution set for |2x – 3| < 7. Its endpoints are the solutions of the corresponding boundary equation |2x – 3| = 7. The solution set for the boundary equation |2x – 3| = 7 is {–2, 5}. The remainder of the real number line must be the solution set for the “greater than” inequality. The solution set for |2x – 3| > 7 is .
Compact compound inequalities of the form – k < M < k as mentioned in Fact B are easy to solve. Such an inequality may always be solved using only the three familiar, fundamental, elementary properties of inequalities. That process is summarized in the next post.

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