LEARN · EXPLAIN · REVISE

Solve by comparing distances to 0 and 2.

Read the idea, work independently, then explain what changed.

2027 JM01 考試大綱 · 7. Algebraic inequalities · PDF 2 / printed page 2

TOPIC 01

Absolute values and solution sets

Build understanding of absolute values and solution sets through definitions, contrasting cases and justified applications.

Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Transfer#Worked example

|x|<|x-2|

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Interpret absolute value as distance and use intersections or unions correctly.

Hint 2

Both sides are nonnegative, so squaring is reversible here.

Worked solution

x<2/2.

Checks and common pitfalls: The midpoint is excluded because the comparison is strict.

Reasoning checklist · self / teacher assessment

Think first. Reveal a hint when the class is ready.

Teacher preparation and assessment

What must be true before using the main rule for absolute values and solution sets?
Which representation makes this task easier, and why?
Change one assumption. Does the conclusion survive?

Choose an equivalent method and explain exclusions before calculating.
Distance: |x−a| measures the distance between x and a on the real line.
Logic of constraints: Simultaneous constraints use intersection; alternative cases use union.
Close with: conditions → representation → reasoning → check.

Expected reasoning: |x−a| measures the distance between x and a on the real line.
Expected reasoning: Simultaneous constraints use intersection; alternative cases use union.
Expected correction: A candidate produced by algebra may violate the original domain or sign condition.