lesson
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If a thermostat is set to 70∘F with an allowable swing of at most 3∘F, your room temperature T must satisfy ∣T−70∣≤3.
The absolute value of a quantity measures its distance from zero on a number line, regardless of direction.
📊Interactive comparison diagram showing absolute value as distance on a horizontal number line from -6 to 6. Top panel: '|x| < 3' highlighted as the bounded interval between -3 and 3 (distance less than 3 from 0). Bottom panel: '|x| > 3' highlighted as two separate outward rays stretching to -infinity and +infinity (distance greater than 3 from 0). Clean labels pointing to the center at 0 and showing distance arrows.
What rule helps us break these absolute value bars down into standard linear inequalities every single time?
The "Less Than" Case: Conjunction (AND)
When an absolute value is isolated and less than a positive number k, the expression is trapped within k units of zero: ∣u∣<k becomes −k<u<k.
This creates an intersection (an AND compound inequality) where both boundaries must be satisfied simultaneously.
📊Step-by-step visual solution breakdown card for '|2x - 3| <= 7'. Step 1: Remove bars and form sandwich inequality '-7 <= 2x - 3 <= 7'. Step 2: Add 3 to all three parts to get '-4 <= 2x <= 10'. Step 3: Divide everything by 2 to reach '-2 <= x <= 5'. Bottom: Number line from -4 to 7 showing closed filled dots at -2 and 5 connected by a solid teal bar.
Let's solve ∣2x−3∣≤7 algebraically by setting up −7≤2x−3≤7.
Adding 3 across all parts gives −4≤2x≤10, and dividing by 2 gives our final bounded solution: [−2,5] or −2≤x≤5.
What happens when the inequality points the opposite direction, demanding a distance larger than a target?
The "Greater Than" Case: Disjunction (OR)
When an absolute value is greater than a positive value k, the quantity lies far out in the tails: ∣u∣>k splits into u<−k or u>k.