A compound inequality combines conditions. AND keeps the overlap shared by every condition; OR keeps every value accepted by at least one condition. Absolute value adds a distance interpretation that naturally produces inside or outside regions.
Intersection versus union
Logical connectors
AND: intersection
A value must satisfy both statements. Solutions often form one bounded interval.
OR: union
A value may satisfy either statement. Solutions may form two separated rays.
Solve an AND inequality
Solve \(7<3x+1\le19\).
- Subtract one from all three parts
\(6<3x\le18\).
- Divide all parts by three
\(2<x\le6\).
- Read the intersection
The left endpoint is open; the right endpoint is closed.
Solve an OR inequality
Solve \(2x-5<-11\) or \(4x+1\ge17\).
- First branch
\(2x<-6\), so \(x<-3\).
- Second branch
\(4x\ge16\), so \(x\ge4\).
- Join with OR
Keep both rays: \(x<-3\) or \(x\ge4\).
Absolute value as distance
The expression \(|E|\) measures the distance of \(E\) from zero. A small-distance condition keeps values between two boundaries; a large-distance condition keeps values outside them.
| Distance condition | Equivalent compound inequality | Geometry |
|---|---|---|
| \(|E|<a\) | \(-a<E<a\) | strictly between |
| \(|E|\le a\) | \(-a\le E\le a\) | between, including endpoints |
| \(|E|>a\) | \(E<-a\) or \(E>a\) | strictly outside |
| \(|E|\ge a\) | \(E\le-a\) or \(E\ge a\) | outside, including endpoints |
Inside-distance case
Solve \(|2x-1|\le7\).
- Write an AND chain
\(-7\le2x-1\le7\).
- Add one throughout
\(-6\le2x\le8\).
- Divide throughout by two
\(-3\le x\le4\).
Outside-distance case
Solve \(|3x+2|>8\).
- Split with OR
\(3x+2<-8\) or \(3x+2>8\).
- Left branch
\(3x<-10\), so \(x<-10/3\).
- Right branch
\(3x>6\), so \(x>2\).
Zero and negative targets
- Because absolute value is nonnegative, \(|E|<0\) has no solution.
- For any positive target, \(|E|>\text{negative target}\) is true for every real input.
- The condition \(|E|\le0\) forces \(E=0\).
- The condition \(|E|\ge0\) is true for every real input.
Integer constraints and extreme values
After solving over the reals, filter the solution set through any integer requirement. For \(-2<x\le4\), the integers are \(-1,0,1,2,3,4\); the least is \(-1\), and the greatest is \(4\).
Check your understanding
Which compound inequality is equivalent to \(|x+1|\ge5\)?
- \(-6\le x\le4\)
- \(x\le-6\) or \(x\ge4\)
- \(-4\le x\le6\)
- \(x<-4\) or \(x>6\)
Show answer and explanation
Answer: \(x\le-6\) or \(x\ge4\)
Distance at least \(5\) means outside or on the boundaries: \(x+1\le-5\) or \(x+1\ge5\).
What to remember
- AND means intersection; OR means union.
- Chained inequalities require applying each operation to all three parts.
- Absolute value less than a positive target describes an inside interval.
- Absolute value greater than a positive target describes two outside rays.
- Inspect zero and negative targets before using a standard pattern.
Put these notes into practice
Apply the ideas with SAT-style questions, then reinforce key details with flashcards.