To solve the absolute value inequality $$|2x - 3| \geq 5$$, follow the five-step systematic procedure for 'greater than' inequalities. Step 1 is to verify the absolute value expression is isolated, which it already is. Step 2 requires writing the equivalent compound inequality: $$2x - 3 \leq -5$$ or $$2x - 3 \geq 5$$. Step 3 involves independently solving these generated inequalities. Adding $$3$$ to both sides of both expressions reduces them to $$2x \leq -2$$ or $$2x \geq 8$$, and subsequently dividing each by $$2$$ yields the final algebraic solution of $$x \leq -1$$ or $$x \geq 4$$. Step 4 is to accurately graph the solution on a standard number line, physically represented by a closed circle on $$-1$$ with shading stretching to the left, alongside a completely separate set starting with a closed circle on $$4$$ continuing rightward. Step 5 dictates expressing this dual, non-overlapping solution correctly utilizing formal interval notation, recorded as $$(-\infty, -1] \cup [4, \infty)$$.

Example: Solving $$|2x - 3| \geq 5$$

To actively practice solving more complex absolute value inequalities incorporating the 'greater than or equal to' property, carefully evaluate the mathematical expressions $$|4x - 3| \geq 5$$ and $$|3x - 4| \geq 2$$. For the inequality $$|4x - 3| \geq 5$$, apply the formal algebraic procedure and translate it entirely into the equivalent compound sequence $$4x - 3 \leq -5$$ or $$4x - 3 \geq 5$$. Resolving for $$x$$ consistently yields the divided parameters $$x \leq -\frac{1}{2}$$ or $$x \geq 2$$, officially recorded utilizing standard interval notation as $$(-\infty, -\frac{1}{2}] \cup [2, \infty)$$. Likewise, the related absolute value inequality $$|3x - 4| \geq 2$$ accurately maps to the equivalent compound equations $$3x - 4 \leq -2$$ or $$3x - 4 \geq 2$$; simplifying logically isolates the specific ranges $$x \leq \frac{2}{3}$$ or $$x \geq 2$$, forming the complete interval notation $$(-\infty, \frac{2}{3}] \cup [2, \infty)$$.

Try It: Solving $$|4x - 3| \geq 5$$ and $$|3x - 4| \geq 2$$

To solve an absolute value inequality featuring a 'greater than' or 'greater than or equal to' symbol, follow a systematic sequential procedure. First, verify that the absolute value expression is completely isolated on one side of the inequality. Second, translate the isolated absolute value inequality into its equivalent compound inequality separated by 'or' (for example, breaking $$|u| \geq a$$ into $$u \leq -a$$ or $$u \geq a$$). Third, fully solve the generated compound inequality to establish the permitted boundary values for the variable. Fourth, accurately graph the entire solution region on a standard number line, extending outward from the endpoints. Finally, properly document the overall mathematical solution utilizing standard interval notation.

Claude

For any algebraic expression $$u$$ and any positive real number $$a$$, the absolute value inequality $$|u| > a$$ translates into the equivalent compound inequality $$u < -a$$ or $$u > a$$. Similarly, the non-strict inequality $$|u| \geq a$$ translates to $$u \leq -a$$ or $$u \geq a$$. This property dictates that the mathematical solution set consists of all values whose calculated distance from zero is strictly greater than (or greater than or equal to) $$a$$. Geometrically, this solution corresponds precisely to the two distinct, non-overlapping regions on the number line extending infinitely outward in opposite directions from the boundaries $$-a$$ and $$a$$.

Absolute Value Inequalities with $$>$$ or $$\geq$$

https://openstax.org/details/books/intermediate-algebra-2e

OpenStax Intermediate Algebra

Consider the introductory absolute value inequality $$|x| \geq 5$$. This mathematical statement describes all numerical values whose distance from zero on the number line is greater than or equal to $$5$$. Because the distinct numbers $$-5$$ and $$5$$ sit exactly five units away from zero, they are both included in the valid solution. Furthermore, any numbers whose distances from zero are greater than five units must be located to the left of $$-5$$ or to the right of $$5$$. Therefore, interpreting this geometric relationship translates the generic inequality $$|x| \geq 5$$ into the corresponding algebraic compound inequality $$x \leq -5$$ or $$x \geq 5$$.

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