Solving 'Greater Than' Absolute-Value Inequalities
Solve 'greater than' absolute-value inequalities in Grade 9 Algebra by splitting into two OR compound inequalities. Master flipping the sign for the negative case every time.
Key Concepts
Property For an inequality in the form $|K| a$, where $K$ represents a variable expression and $a 0$, the solution is $K < a$ OR $K a$. Explanation This rule is for when your value is far from zero. Your distance must be greater than a certain number, which means you can either be far out in the positive numbers OR far out in the negative numbers. Since you can't be in both places at once, the solution splits into two separate regions. Examples $|x + 7| 3$ is solved as $x + 7 < 3$ OR $x + 7 3$, giving the solution $x < 10$ OR $x 4$. $ 2|x| < 6$ simplifies to $|x| 3$ (remember to flip the sign!), so the solution is $x < 3$ OR $x 3$.
Common Questions
How do you solve a 'greater than' absolute-value inequality?
Apply the OR rule: split |K| > a into two inequalities, K > a and K < -a. Solve each separately and combine the solutions with OR. The negative case requires flipping the inequality sign.
Why does the negative case flip the inequality sign in absolute-value inequalities?
When you apply the negative case of |K| > a, you write K < -a because values to the left of -a are more than 'a' units from zero. The direction of the inequality reverses because you are comparing to a negative bound.
What is the 'great-OR' memory trick for absolute-value inequalities?
The word 'greater' contains 'OR,' reminding you that |K| > a always produces an OR compound inequality with two separate solution regions. When the inequality is less than, it produces an AND condition instead.