Absolute Value Inequalities
Solve absolute value inequalities in Grade 10 algebra. Apply the rules |x|<a gives -a<x<a and |x|>a gives x>a or x<-a to solve and graph compound inequality solutions.
Key Concepts
For any positive number $a$: $|x| < a$ is equivalent to the conjunction $ a < x < a$. $|x| a$ is equivalent to the disjunction $x a$ or $x < a$.
$|x 2| < 5$ is a conjunction that becomes $ 5 < x 2 < 5$, which simplifies to $ 3 < x < 7$. $|x + 4| \geq 6$ is a disjunction that becomes $x + 4 \geq 6$ or $x + 4 \leq 6$, which simplifies to $x \geq 2$ or $x \leq 10$.
Think of it this way: 'less than' keeps the solutions together, 'AND wiched' between two endpoints. 'Greater than' splits the solutions apart, sending them in opposite directions with an 'OR'. Use the memory tricks 'less th AND' for conjunctions and 'great OR' for disjunctions to remember if your solution graph will be a single segment or two separate rays.
Common Questions
What are the two cases for solving absolute value inequalities?
For |x| < a (positive a): solve as conjunction -a < x < a. For |x| > a: solve as disjunction x > a or x < -a.
How do you solve |2x - 3| ≤ 7?
Set up -7 ≤ 2x - 3 ≤ 7. Add 3: -4 ≤ 2x ≤ 10. Divide by 2: -2 ≤ x ≤ 5. The solution is the closed interval [-2, 5].
How do you graph absolute value inequality solutions on a number line?
For conjunction solutions, shade the segment between two boundary points. For disjunction, shade two rays pointing outward from the boundaries. Use closed/open circles for inclusive/exclusive endpoints.