Compound Inequalities: When to Use AND or OR

A compound inequality can have two inequality signs, as in 2 < x ≤ 7, or two separate statements joined by the words and or or. The arithmetic is usually straightforward. The key is translating the relationship correctly: must a value meet both limits, or is meeting either one enough? That choice determines whether the solution is an overlap or a union.

AND means both conditions must be true

Consider x > 2 and x < 7. A solution must be greater than 2 and less than 7 at the same time. The values that meet both are between 2 and 7, so the solution is 2 < x < 7. On a number line, shade to the right of 2 for the first condition and to the left of 7 for the second. Keep only the region where the two shadings overlap.

The same set can be written in interval notation as (2, 7). Parentheses mean the endpoints are excluded because the inequalities are strict. If the problem instead says x ≥ 2 and x < 7, the interval is [2, 7): include 2, exclude 7. Brackets and parentheses encode whether each boundary belongs to the solution.

Solve a three-part inequality together

A statement such as 3 ≤ 2x + 1 < 11 means both 3 ≤ 2x + 1 and 2x + 1 < 11. You can preserve the chain while performing the same operation on all three parts. Subtract 1 throughout to get 2 ≤ 2x < 10. Divide every part by 2 to obtain 1 ≤ x < 5. The solution interval is [1, 5).

Treat the chain as two inequalities joined by AND. If you prefer, solve each separately: 3 ≤ 2x + 1 gives x ≥ 1, and 2x + 1 < 11 gives x < 5. Their intersection is again 1 ≤ x < 5. Applying the same operation to all three expressions is a compact method, but it still depends on the usual inequality rule.

OR means either condition is enough

Now consider x < 1 or x > 4. A value is a solution if it lies below 1 or above 4. It does not need to satisfy both. The solution is the two outer regions, written (−∞, 1) ∪ (4, ∞). On the number line, shade left of 1 and right of 4; keep both regions because the statements are joined by OR.

An OR solution may be one continuous region if its component sets overlap or touch in a way that covers all points between them. For example, x < 5 or x > 2 is true for every real number: any number is either below 5 or above 2. Testing a few values can reveal this, but interval reasoning confirms that the union covers the full number line.

Translate everyday descriptions

Words such as between often describe AND. A temperature above 60 degrees and below 80 degrees is 60 < t < 80. A condition that allows an amount up to 80 and no lower than 60 becomes 60 ≤ t ≤ 80. The exact boundary depends on words such as at least, more than, no greater than, or under.

The phrase outside a range often describes OR. A value more than 4 units away from zero means x < −4 or x > 4. A membership rule saying a person must be younger than 18 or at least 65 also uses OR: either age range qualifies. Translate the words before doing algebra, especially when an endpoint is included in one condition but not the other.

Worked example: a distance condition

A delivery route is considered within the target band when its distance d is no more than 3 miles from a planned distance of 12 miles. “No more than 3 miles away” means |d − 12| ≤ 3. This absolute-value statement can be expanded as −3 ≤ d − 12 ≤ 3, an AND compound inequality. Add 12 throughout: 9 ≤ d ≤ 15. Distances from 9 through 15 miles, inclusive, satisfy the condition.

For distances more than 3 miles away, the inequality is |d − 12| > 3. This expands to d − 12 < −3 or d − 12 > 3, so d < 9 or d > 15. It is an OR because a distance can be too low or too high. The absolute-value symbol describes distance from a center; the inequality sign determines whether the solution is between the boundaries or outside them.

When solving, watch for a negative multiplier

The rule for inequalities still applies inside each compound statement: multiplying or dividing by a negative number reverses the inequality sign. Suppose −6 < −2x ≤ 4. Divide each part by −2 and reverse both inequality signs: 3 > x ≥ −2. Rewrite in increasing order as −2 ≤ x < 3. Interval notation is [−2, 3). A common error is to reverse only one sign or forget to reorder the endpoints.

For two separate statements joined by OR, solve each inequality independently, reverse a sign whenever an operation requires it, and then combine their solution sets. For AND, solve both and find their overlap. Keeping the logic separate from the algebra makes it easier to catch errors.

Check the result with test values

Choose one value that should be inside each proposed interval and substitute it into the original statements. For 1 ≤ x < 5, test x = 2: both 1 ≤ 2 and 2 < 5 are true. Test an outside value such as x = 6: the upper condition fails, so 6 is excluded. For x < 1 or x > 4, test x = 0 and x = 5; each satisfies at least one branch. A value such as 2 satisfies neither and should remain unshaded.

Boundary tests help too. If the original symbol is ≤, the boundary is included; if it is <, the boundary is excluded. Substitute the endpoint into the original inequality rather than relying only on memory of interval notation. When graphing, use a closed dot for an included endpoint and an open dot for an excluded endpoint.

Frequent errors

• Keeping only the overlap for an OR statement. OR means the union, so either region is included.
• Keeping both outer regions for an AND statement. AND requires both conditions, so only the intersection works.
• Including an endpoint from a strict inequality or excluding it from an inclusive inequality.
• Reversing an inequality sign when adding or subtracting a number. Only multiplication or division by a negative number reverses it.
• Dividing a chained inequality by a negative value without reversing every inequality sign in the chain.
• Using a decimal approximation or a rough sketch to decide an exact boundary that can be found algebraically.
• Solving the expressions correctly but forgetting to interpret the result as a set satisfying the original conditions.

A dependable method

1. Underline the connector: AND or OR. Rewrite a word problem as two precise conditions if needed.
2. Solve a three-part chain on all sides at once, or split it into two inequalities joined by AND.
3. For separate inequalities, solve each one and reverse the sign only when multiplying or dividing by a negative number.
4. Combine the results using intersection for AND and union for OR.
5. Mark open or closed endpoints from the original inequality signs, then write interval notation if requested.
6. Test an interior value and any important boundary in the original statement.

Exam takeaway

AND keeps values that satisfy both inequalities; OR keeps values that satisfy at least one. Read the connector first, solve carefully, and then combine the intervals as an intersection or union. On a graph, overlap is the AND solution, while every region included by either branch belongs to the OR solution.