The Central Idea
Some inequalities describe one boundary:
\[ x<5. \]Others describe two boundaries at once:
\[ 2These are compound inequalities.
Absolute-value inequalities often produce exactly these two kinds of solution sets:
- inside a distance produces an interval;
- outside a distance produces two rays.
What Is a Compound Inequality?
A compound inequality combines two or more inequalities using logical words such as and or or.
For example,
\[ x>2 \quad\text{and}\quad x<7 \]means that both conditions must be true.
This can be written compactly as
\[ 2means that either condition is sufficient.
“And” Means Intersection
For an and compound inequality, a number must satisfy both conditions.
Consider
\[ x\ge1 \quad\text{and}\quad x<6. \]The numbers satisfying both are those from \(1\) up to but not including \(6\):
\[ 1\le x<6. \]In interval notation:
\[ [1,6). \]This shared region is the intersection of the two solution sets.
“Or” Means Union
For an or compound inequality, a number may satisfy either condition.
Consider
\[ x\le-2 \quad\text{or}\quad x>4. \]The solution set contains two separate regions:
\[ (-\infty,-2] \cup (4,\infty). \]The symbol
\[ \cup \]denotes the union of the two sets.
Chained Inequalities
An expression such as
\[ 3Its interval notation is
\[ (3,8]. \]Solving a Three-Part Inequality
Consider
\[ 2<3x+5<17. \]We can perform the same operation on all three parts.
Subtract \(5\):
\[ -3<3x<12. \]Divide by \(3\):
\[ -1Negative Division in a Chained Inequality
Consider
\[ -6<-2x\le10. \]Divide all three parts by \(-2\).
Because the divisor is negative, both inequality directions reverse:
\[ 3>x\ge-5. \]Rewrite in increasing order:
\[ -5\le x<3. \]Interval notation:
\[ [-5,3). \]Why Both Directions Reverse
In
\[ aThe order becomes reversed:
\[ -ac<-bc<-aa \]when written from smallest to largest.
So every inequality symbol affected by multiplication or division by a negative quantity must reverse.
Solving an “And” Compound Inequality Separately
Consider
\[ 2x-1\ge5 \quad\text{and}\quad 3x+2<14. \]Solve the first:
\[ 2x\ge6, \] \[ x\ge3. \]Solve the second:
\[ 3x<12, \] \[ x<4. \]Both must hold, so
\[ 3\le x<4. \]Interval notation:
\[ [3,4). \]Solving an “Or” Compound Inequality Separately
Consider
\[ 2x+1<-5 \quad\text{or}\quad 3x-2\ge10. \]First branch:
\[ 2x<-6, \] \[ x<-3. \]Second branch:
\[ 3x\ge12, \] \[ x\ge4. \]Therefore,
\[ x<-3 \quad\text{or}\quad x\ge4. \]Interval notation:
\[ (-\infty,-3) \cup [4,\infty). \]When an “And” Compound Has No Solution
Consider
\[ x>5 \quad\text{and}\quad x<2. \]No real number can satisfy both conditions.
Therefore,
\[ \varnothing \]is the solution set.
When an “Or” Compound Gives All Real Numbers
Consider
\[ x<5 \quad\text{or}\quad x\ge2. \]Every real number satisfies at least one of these conditions.
Therefore, the union is
\[ \mathbb{R}. \]Absolute Value as Distance
The expression
\[ |x-a| \]is the distance between \(x\) and \(a\) on the real number line.
This gives a geometric meaning to absolute-value inequalities.
For example,
\[ |x-3|<5 \]means:
The distance from \(x\) to \(3\) is less than \(5\).
So \(x\) must lie within 5 units of \(3\).
Less Than Means Inside
Consider
\[ |x|<4. \]The numbers whose distance from zero is less than \(4\) lie between \(-4\) and \(4\):
\[ -4Less Than or Equal Means Inside Including the Boundary
Consider
\[ |x|\le4. \]The distance from zero may be at most \(4\).
Therefore,
\[ -4\le x\le4. \]Interval notation:
\[ [-4,4]. \]Greater Than Means Outside
Consider
\[ |x|>4. \]The distance from zero must exceed \(4\).
So \(x\) lies outside the interval from \(-4\) to \(4\):
\[ x<-4 \quad\text{or}\quad x>4. \]Thus,
\[ |x|>4 \quad\Longleftrightarrow\quad x<-4 \ \text{or}\ x>4. \]Greater Than or Equal Means Outside Including the Boundary
For
\[ |x|\ge4, \]we include the boundary points:
\[ x\le-4 \quad\text{or}\quad x\ge4. \]Interval notation:
\[ (-\infty,-4] \cup [4,\infty). \]The Four Core Absolute-Value Patterns
For \(k>0\):
| Absolute-Value Form | Equivalent Compound Inequality |
|---|---|
\(|A|| \(-k | |
| \(|A|\le k\) | \(-k\le A\le k\) |
| \(|A|>k\) | \(A<-k\) or \(A>k\) |
| \(|A|\ge k\) | \(A\le-k\) or \(A\ge k\) |
A Memory Aid That Reflects the Geometry
For positive \(k\):
- less than means inside the two boundary points;
- greater than means outside the two boundary points.
This is more reliable than memorizing isolated sign patterns because it comes directly from distance.
Shifted Absolute-Value Inequality
Solve
\[ |x-3|<5. \]Use the inside pattern:
\[ -5These are exactly the numbers within 5 units of \(3\).
Shifted Outside Inequality
Solve
\[ |x+2|\ge6. \]Use the outside pattern:
\[ x+2\le-6 \quad\text{or}\quad x+2\ge6. \]Subtract \(2\):
\[ x\le-8 \quad\text{or}\quad x\ge4. \]Interval notation:
\[ (-\infty,-8] \cup [4,\infty). \]Isolate the Absolute Value First
Consider
\[ 2|x-1|+3<11. \]Subtract \(3\):
\[ 2|x-1|<8. \]Divide by \(2\):
\[ |x-1|<4. \]Now convert to a compound inequality:
\[ -4Multistep Outside Example
Solve
\[ 3|2x+1|-6\ge9. \]Add \(6\):
\[ 3|2x+1|\ge15. \]Divide by \(3\):
\[ |2x+1|\ge5. \]Split into outside branches:
\[ 2x+1\le-5 \]or
\[ 2x+1\ge5. \]First branch:
\[ 2x\le-6, \] \[ x\le-3. \]Second branch:
\[ 2x\ge4, \] \[ x\ge2. \]Therefore,
\[ (-\infty,-3] \cup [2,\infty). \]Absolute Value with a Negative Right Side
Because absolute value is always nonnegative,
\[ |A|with negative \(k\) has no real solution.
For example,
\[ |x-2|<-1 \]has solution set
\[ \varnothing. \]Greater Than a Negative Number
Now consider
\[ |x-2|>-1. \]Since
\[ |x-2|\ge0 \]for every real \(x\), it is automatically true that
\[ |x-2|>-1. \]Therefore, the solution set is
\[ \mathbb{R}. \]The Zero Boundary Case
Consider
\[ |x-4|<0. \]No absolute value can be negative, so there is no solution:
\[ \varnothing. \]But
\[ |x-4|\le0 \]forces
\[ |x-4|=0, \]so
\[ x=4. \]Greater Than Zero
Consider
\[ |x-4|>0. \]Absolute value equals zero only when
\[ x-4=0. \]So every real number except \(4\) satisfies the inequality:
\[ x<4 \quad\text{or}\quad x>4. \]Interval notation:
\[ (-\infty,4) \cup (4,\infty). \]Absolute Value and Center-Radius Form
The inequality
\[ |x-a|\le r \]with \(r\ge0\) describes the closed interval centered at \(a\) with radius \(r\):
\[ a-r\le x\le a+r. \]Likewise,
\[ |x-a|Converting an Interval into Absolute-Value Form
Suppose
\[ 2The distance from the midpoint to either endpoint is
\[ 4. \]Therefore,
\[ 2Closed Interval to Absolute Value
Consider
\[ -3\le x\le7. \]The midpoint is
\[ \frac{-3+7}{2}=2. \]The radius is
\[ 5. \]So the interval is equivalent to
\[ |x-2|\le5. \]Outside an Interval in Absolute-Value Form
Suppose
\[ x<-1 \quad\text{or}\quad x>9. \]The midpoint between the boundary points is
\[ 4, \]and the distance to either boundary is
\[ 5. \]So the condition can be written as
\[ |x-4|>5. \]Testing a Compound Solution
Suppose we obtain
\[ -2Then test the boundary points:
- \(-2\) should be excluded;
- \(5\) should be included.
Boundary testing is especially useful for catching strict-versus-inclusive errors.
Testing an “Or” Solution
Suppose the solution is
\[ x<-3 \quad\text{or}\quad x\ge4. \]Useful test values include:
- a value in the left region, such as \(-5\);
- a value in the gap, such as \(0\);
- a value in the right region, such as \(5\);
- the boundary values \(-3\) and \(4\).
This checks both rays, the excluded middle region, and endpoint behavior.
Application: Tolerance
Suppose a machine part should be \(20\) mm long with a tolerance of at most \(0.1\) mm.
If \(x\) is the actual length, then
\[ |x-20|\le0.1. \]Convert to a compound inequality:
\[ -0.1\le x-20\le0.1. \]Add \(20\):
\[ 19.9\le x\le20.1. \]The acceptable lengths form a closed interval.
Application: Distance from a Target
Suppose a temperature must differ from \(72^\circ\) by more than \(5^\circ\).
If \(T\) is the temperature, then
\[ |T-72|>5. \]Therefore,
\[ T<67 \]or
\[ T>77. \]This produces two separate allowable regions.
Application: Range Requirement
Suppose a score must be at least \(70\) but no more than \(90\).
If \(s\) is the score, then
\[ 70\le s\le90. \]This is an and compound inequality because both bounds must be satisfied.
Worked Example: Chained Inequality
Solve
\[ 5\le2x+1<13. \]Subtract \(1\):
\[ 4\le2x<12. \]Divide by \(2\):
\[ 2\le x<6. \]Interval notation:
\[ [2,6). \]Worked Example: Chained Inequality with Negative Division
Solve
\[ -12< -3x \le6. \]Divide all parts by \(-3\) and reverse both directions:
\[ 4>x\ge-2. \]Rewrite:
\[ -2\le x<4. \]Interval notation:
\[ [-2,4). \]Worked Example: Absolute Value Inside
Solve
\[ |2x-3|\le7. \]Write the compound inequality:
\[ -7\le2x-3\le7. \]Add \(3\):
\[ -4\le2x\le10. \]Divide by \(2\):
\[ -2\le x\le5. \]Therefore,
\[ [-2,5]. \]Worked Example: Absolute Value Outside
Solve
\[ |3x+2|>8. \]Split:
\[ 3x+2<-8 \]or
\[ 3x+2>8. \]First branch:
\[ 3x<-10, \] \[ x<-\frac{10}{3}. \]Second branch:
\[ 3x>6, \] \[ x>2. \]Therefore,
\[ \left(-\infty,-\frac{10}{3}\right) \cup (2,\infty). \]A Reliable Strategy for Compound Inequalities
- Identify whether the connector is and or or.
- Solve each inequality while preserving order.
- Reverse inequality directions when multiplying or dividing by a negative quantity.
- For and, keep only the overlap.
- For or, combine all allowed regions.
- Express the result with inequality notation, interval notation, or a number-line graph.
A Reliable Strategy for Absolute-Value Inequalities
- Isolate the absolute-value expression.
- Inspect the comparison value.
- If it is positive, use the inside or outside pattern.
- If it is zero or negative, analyze the special case directly.
- Solve the resulting compound inequality or branches.
- Write the final solution set clearly.
- Check boundaries and representative interior or exterior values when useful.
Common Errors
Confusing “And” with “Or”
And means both conditions must hold. Or means either condition is enough.
Using the Wrong Absolute-Value Pattern
For positive \(k\):
\[ |A|means outside.
Forgetting to Isolate the Absolute Value
Do not convert to a compound inequality until the absolute-value expression stands alone.
Forgetting to Reverse Both Signs in a Chained Inequality
Dividing all three parts by a negative number reverses both inequality directions.
Using Intersection for an “Or” Statement
An “or” compound inequality uses union, not overlap.
Ignoring Special Cases When the Comparison Value Is Nonpositive
Absolute value is never negative, so cases involving \(0\) or negative bounds should be analyzed before applying memorized formulas.
Practice
1. Chained Inequality
Solve:
\[ 1<2x+3\le9. \]Check Answer
\[ -2<2x\le6, \]so
\[ -12. “And” Compound
Solve:
\[ x\ge2 \quad\text{and}\quad x<7. \]Check Answer
\[ 2\le x<7. \]Interval notation:
\[ [2,7). \]3. “Or” Compound
Solve:
\[ x<-4 \quad\text{or}\quad x\ge1. \]Check Answer
\[ (-\infty,-4) \cup [1,\infty). \]4. Absolute Value: Inside
Solve:
\[ |x-2|<6. \]Check Answer
\[ -65. Absolute Value: Outside
Solve:
\[ |x+3|\ge5. \]Check Answer
\[ x+3\le-5 \]or
\[ x+3\ge5. \]So
\[ x\le-8 \]or
\[ x\ge2. \]6. Multistep Absolute Value
Solve:
\[ 2|x-1|+4\le10. \]Check Answer
\[ 2|x-1|\le6, \] \[ |x-1|\le3. \]Therefore,
\[ -3\le x-1\le3, \]so
\[ -2\le x\le4. \]7. Special Case
Solve:
\[ |x+5|<-2. \]Check Answer
No real number has a negative absolute value, so
\[ \varnothing. \]Key Ideas
- Compound inequalities combine multiple conditions using logical connectors such as and and or.
- And corresponds to intersection: every condition must hold.
- Or corresponds to union: at least one condition must hold.
- Chained inequalities are compact forms of and statements.
- When a chained inequality is multiplied or divided by a negative quantity, every affected inequality direction reverses.
- Absolute value measures distance from a center.
- For positive bounds, absolute-value inequalities with less than describe inside intervals.
- For positive bounds, absolute-value inequalities with greater than describe outside regions.
- Strict inequalities produce open boundaries; inclusive inequalities produce closed boundaries.
- Absolute-value inequalities with zero or negative comparison values require special-case reasoning.
- Intervals can often be rewritten in center-radius absolute-value form.
- Compound and absolute-value inequalities model ranges, tolerances, minimums, maximums, and distance constraints.
Where This Leads
Compound and absolute-value inequalities show that solution sets may be single intervals, unions of intervals, all real numbers, or the empty set.
This prepares the way for polynomial and rational inequalities, sign charts, systems of inequalities, feasible regions, optimization, and more advanced constraint-based reasoning.