Compound Inequalities

A compound inequality joins two inequalities into a single statement using the word and or or. Instead of finding one boundary, you find two — and the word connecting them tells you whether the solution is the overlap of both ranges or the combination of both ranges.

”And” Compound Inequalities (Intersection)

An “and” compound inequality requires both conditions to be true simultaneously. The solution is the intersection — only the values that satisfy both inequalities at the same time.

You will most often see “and” compound inequalities written in combined form:

$a < x < b$

This compact notation means $x > a$ and $x < b$. The variable sits between two boundaries, and you solve by performing the same operation on all three parts.

Example 1: Solve $-3 < 2x + 1 \leq 7$

This is an “and” compound inequality in combined form. Work with all three parts at once.

Step 1 — Subtract 1 from all three parts:

$-3 - 1 < 2x \leq 7 - 1$

$-4 < 2x \leq 6$

Step 2 — Divide all three parts by 2:

$-2 < x \leq 3$

Answer: $x$ is greater than $-2$ (not including $-2$) and less than or equal to $3$ (including $3$).

Number line: Open circle at $-2$, closed circle at $3$, shading between them.

Interval notation: $(-2, 3]$

Example 2: Solve $4 \leq 3x - 2$ and $3x - 2 < 13$

When given as two separate statements, you can combine them first or solve each one independently.

Solve each part:

First inequality: $4 \leq 3x - 2 \implies 6 \leq 3x \implies 2 \leq x$

Second inequality: $3x - 2 < 13 \implies 3x < 15 \implies x < 5$

Combine the results: $2 \leq x$ and $x < 5$, which gives $2 \leq x < 5$.

Interval notation: $[2, 5)$

When “And” Produces No Solution

If the two conditions don’t overlap, the compound inequality has no solution.

For example, solve $x > 7$ and $x < 3$.

No number is simultaneously greater than 7 and less than 3. The solution set is empty — written as $\emptyset$ or “no solution."

"Or” Compound Inequalities (Union)

An “or” compound inequality requires at least one condition to be true. The solution is the union — all values that satisfy either inequality (or both).

“Or” compound inequalities typically produce two separate regions on the number line with a gap between them.

Example 3: Solve $5x - 1 < -11$ or $5x - 1 > 14$

Solve each inequality separately:

First: $5x - 1 < -11 \implies 5x < -10 \implies x < -2$

Second: $5x - 1 > 14 \implies 5x > 15 \implies x > 3$

Answer: $x < -2$ or $x > 3$.

Number line: Open circle at $-2$ with an arrow pointing left, plus open circle at $3$ with an arrow pointing right. The region between $-2$ and $3$ is not shaded.

Interval notation: $(-\infty, -2) \cup (3, \infty)$

The $\cup$ symbol means “union” — the combination of both sets.

Example 4: Solve $-2x + 3 \geq 9$ or $x - 4 \geq 1$

Solve each:

First: $-2x + 3 \geq 9 \implies -2x \geq 6 \implies x \leq -3$ (flip the sign — dividing by $-2$)

Second: $x - 4 \geq 1 \implies x \geq 5$

Answer: $x \leq -3$ or $x \geq 5$.

Interval notation: $(-\infty, -3] \cup [5, \infty)$

When “Or” Produces All Real Numbers

If the two regions overlap to cover the entire number line, the solution is all real numbers.

For example: $x > 1$ or $x < 4$.

Every real number satisfies at least one of these conditions (and most satisfy both), so the solution is all real numbers: $(-\infty, \infty)$.

Interval Notation Summary

Interval notation is a compact way to express solution sets. Here is a reference:

• Parentheses $( \;)$ mean the endpoint is not included (strict inequality)
• Brackets $[ \;]$ mean the endpoint is included (non-strict inequality)
• Infinity always gets a parenthesis — you can never “reach” infinity

Solving Compound Inequalities with Negative Coefficients

The sign-flip rule from basic inequalities still applies. When you multiply or divide any part of a compound inequality by a negative number, flip all inequality signs in that step.

Example 5: Solve $-15 \leq -3x < 6$

Step 1 — Divide all three parts by $-3$ and flip both inequality signs:

$\frac{-15}{-3} \geq x > \frac{6}{-3}$

$5 \geq x > -2$

Step 2 — Rewrite in standard left-to-right order:

$-2 < x \leq 5$

Interval notation: $(-2, 5]$

Notice how both inequality signs flipped: $\leq$ became $\geq$ and $<$ became $>$. Then we rewrote so the smaller number is on the left.

Real-World Application: HVAC — Thermostat Temperature Range

An HVAC technician is programming a commercial thermostat system for an office building. The building code requires the indoor temperature $T$ (in degrees Fahrenheit) to stay within a comfort range:

$68 \leq T \leq 76$

The thermostat has a calibration offset. After testing, the technician finds the sensor reads 3 degrees higher than the actual temperature. If $S$ is the sensor reading, then the actual temperature is $T = S - 3$. To maintain the required range:

$68 \leq S - 3 \leq 76$

Add 3 to all parts:

$71 \leq S \leq 79$

The technician programs the thermostat to maintain a sensor reading between 71 and 79 degrees. If the sensor reads below 71, the heater turns on. If it reads above 79, the cooling system activates.

This is a classic “and” compound inequality — the temperature must satisfy both the lower bound and the upper bound simultaneously.

Nursing — IV Drip Rate Monitoring

A nurse is monitoring an IV drip that must deliver medication at a rate $R$ (in mL per hour). The doctor’s order specifies:

• The rate must be at least 50 mL/hr to be therapeutic
• The rate must not exceed 80 mL/hr to avoid adverse effects

The constraint is:

$50 \leq R \leq 80$

The IV pump displays the rate in drops per minute. The conversion factor is 15 drops per mL for the tubing being used, so $D = \frac{R \times 15}{60} = \frac{R}{4}$ where $D$ is drops per minute.

Substituting $R = 4D$:

$50 \leq 4D \leq 80$

Divide all parts by 4:

$12.5 \leq D \leq 20$

The nurse monitors the drip to confirm between 12.5 and 20 drops per minute. A rate outside this range signals a problem — too slow means the medication won’t be effective, too fast risks toxicity.

Common Mistakes to Avoid

1. Confusing “and” with “or.” With “and,” the solution is where both regions overlap (smaller region). With “or,” the solution is where either region exists (larger region). Drawing a number line for each helps you see the difference.

2. Forgetting to flip all signs when dividing by a negative. In a three-part compound inequality like $-15 \leq -3x < 6$, both inequality symbols must flip when dividing by $-3$.

3. Writing impossible “and” inequalities without checking. If solving produces $x > 7$ and $x < 3$, don’t write $7 < x < 3$ — that is nonsensical. The answer is no solution.

4. Dropping the “or” when writing interval notation. The solution $x < -2$ or $x > 3$ must be written as $(-\infty, -2) \cup (3, \infty)$. Writing $(-2, 3)$ would mean the opposite — values between $-2$ and $3$.

5. Using brackets with infinity. Always use parentheses with $\infty$ and $-\infty$. Writing $[-\infty, 3]$ is incorrect because infinity is not a number that can be “included.”

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• “And” compound inequalities require both conditions to be true — the solution is the intersection (overlap) of two ranges
• “Or” compound inequalities require at least one condition to be true — the solution is the union (combination) of two ranges
• In combined form ($a < x < b$), perform operations on all three parts simultaneously
• Flip all inequality signs when multiplying or dividing by a negative number — this applies to every sign in the compound inequality
• Interval notation uses parentheses for excluded endpoints and brackets for included endpoints — infinity always gets a parenthesis
• “And” can produce no solution if the regions don’t overlap; “or” can produce all real numbers if the regions cover the entire number line

Return to Algebra for more topics in this section.