Inequalities, intervals and absolute value

An inequality asks for a set of values

An equation asks when two expressions are equal. An inequality compares their order. The statement x < 3 accepts every real number less than 3, including negative numbers and fractions, but excludes 3 itself. Solving an inequality means describing all permitted values that make its original comparison true.

This lesson uses real numbers unless a problem restricts them. The symbols < and > are strict: equality is excluded. The symbols ≤ and ≥ include equality. “At most 3” means x ≤ 3; “at least 3” means x ≥ 3. “Less than 3” and “at most 3” differ precisely at the boundary.

Show membership and direction on a number line

For x < 3, put an open circle at 3 and mark the line to its left. The open circle says the boundary is excluded. For x ≥ 3, use a filled circle at 3 and mark to the right. Arrows indicate continuation beyond the visible drawing; a diagram’s edge is not a final solution value.

A number line represents a continuous set of real values, not only the labelled integers. Check a boundary by substituting it into the original statement: 3 satisfies x ≥ 3 but does not satisfy x > 3. The shading and endpoint marker must agree with that check.

Write the same set compactly as an interval

An interval contains every real number between its stated bounds. Parentheses exclude a finite endpoint; square brackets include it. Thus (−2, 5] means −2 < x ≤ 5. The comma separates the lower and upper bounds; these are not the coordinates of a single point.

Unbounded intervals use −∞ or ∞ with parentheses, such as (−∞, 3) or [3, ∞). Infinity is not a real endpoint that can be included. We use this parenthesis-and-bracket convention throughout Ark; other school traditions may write open intervals with outward-facing square brackets. The empty set, written ∅, contains no values, and (−∞, ∞) is the whole real number line.

Preserve order when the operation preserves order

As with equations, replacing an expression by an identical expression does not change the solution set. Adding or subtracting the same defined expression on both sides also preserves the comparison. From 3x − 1 < 5, add 1 to both sides to obtain 3x < 6, then divide by the positive number 3 to get x < 2.

Adding a negative number still preserves order: from 2 < 5, adding −7 gives −5 < −2. Multiplying or dividing both sides by a fixed positive number preserves order as well. Each of these steps can be undone, so the final inequality describes exactly the original solutions, rather than merely producing some examples.

Reverse the comparison when scaling by a negative number

Multiplication by −1 reflects the number line across zero: the smaller number becomes the larger one. For example, 2 < 5 becomes −2 > −5. Any fixed negative multiplier reverses order; division by a negative number does the same. Strict comparisons remain strict, and inclusive comparisons remain inclusive.

To solve −2x + 5 ≥ 1, subtract 5 to obtain −2x ≥ −4. Dividing by −2 reverses ≥ to ≤, giving x ≤ 2. The boundary 2 gives equality in the original; 1 satisfies it and 3 does not. Multiplication by zero loses information, and division by zero is undefined. Do not multiply or divide by an expression of unknown sign without separate justified cases.

Check constant statements and permitted values

Collecting terms may remove the unknown. The inequality 2x + 1 < 2x + 3 reduces to 1 < 3, true for every real x. In contrast, 2x + 1 ≥ 2x + 3 reduces to 1 ≥ 3, false for every real x. A missing unknown does not automatically mean no solution: inspect the remaining statement.

A context may narrow the real solution set. If n counts whole tickets, 4n + 3 ≤ 18 gives n ≤ 15/4 algebraically. Combining this with nonnegative whole-number counts leaves 0, 1, 2 and 3. These four values are not the full real interval; they are the solutions in the stated ticket domain.

AND keeps overlap; OR keeps either condition

A compound condition joins comparisons. AND requires both to be true. For x > −1 and x ≤ 3, the solutions are their overlap, (−1, 3]. This overlap is called an intersection, denoted ∩. The chain −1 < x ≤ 3 is another way to write this AND condition.

OR is inclusive here: either condition, or both, is enough. For x ≤ −2 or x > 3, keep both rays, written (−∞, −2] ∪ (3, ∞). The symbol ∪ means union. An AND overlap can be empty, as in x < 0 and x ≥ 2. An OR union can cover every real number, as in x < 2 or x ≥ 1.

In a chained inequality, apply an operation to all three parts. From −1 < 2x + 1 ≤ 7, subtract 1 everywhere to get −2 < 2x ≤ 6, then divide every part by 2 to get −1 < x ≤ 3. If dividing a chain by a negative number, reverse both comparison signs and rewrite the bounds in increasing order.

Read absolute value as distance from a point

The quantity |x − a| is the nonnegative distance between x and a on the real number line. For |x − 1| < 2, find points less than 2 units from 1: −1 < x < 3. The two bounds are 1 minus 2 and 1 plus 2, and both are excluded because the distance comparison is strict.

For a positive radius r, a distance less than r gives an inside interval: |x − a| < r means a − r < x < a + r. A distance greater than r gives two outside rays: |x − a| > r means x < a − r OR x > a + r. Use inclusive endpoints for ≤ or ≥. First isolate the absolute-value expression when other operations surround it; its distance comparison must be read after those operations are handled.

Use nonnegativity before applying a template

The familiar inside/outside templates above assume r is positive. A distance cannot be negative. Therefore |x − a| < 0 has no solutions, |x − a| ≤ 0 has only x = a, |x − a| > 0 accepts every real value except a, and |x − a| ≥ 0 accepts every real value.

For a negative threshold r, both |x − a| < r and |x − a| ≤ r are impossible; both |x − a| > r and |x − a| ≥ r hold for every real x. A negative threshold is not a negative geometric radius. These cases follow directly from nonnegativity and should be checked before constructing endpoints.

Compare a complete solution set with individual checks

Choose a condition and a candidate independently in the model. Its number line shows the full symbolic solution set on a fixed scale, while the table checks individual candidates against the original condition. For distance conditions, vary the threshold through positive, zero and negative cases.

Open and filled endpoints state membership; arrows state unbounded continuation. A single-point solution is drawn as one filled point. The finite candidate list can reveal mistakes, but cannot establish the entire solution set by itself. The equivalent algebra or distance argument supplies that justification.

Worked examples

Try it yourself

1. Write −3 ≤ x < 2 in interval notation. Is x = 2 allowed?

   Show the answer and reasoning

   [−3, 2); 2 is excluded.

   The lower comparison includes equality, so −3 gets a square bracket. The upper comparison is strict, so 2 gets a parenthesis and an open marker.

2. Solve 5 − 3x ≤ 11. Explain the comparison direction.

   Show the answer and reasoning

   x ≥ −2, or [−2, ∞).

   Subtracting 5 gives −3x ≤ 6. Dividing by the negative number −3 reverses ≤ to ≥. Adding or subtracting a negative number alone would not reverse it.

3. Solve −4 < 2x − 2 ≤ 6.

   Show the answer and reasoning

   −1 < x ≤ 4, or (−1, 4].

   Add 2 to all three parts to obtain −2 < 2x ≤ 8, then divide all three parts by the positive number 2. Only the upper endpoint is included.

4. Compare x < 0 AND x ≥ 2 with x < 0 OR x ≥ 2.

   Show the answer and reasoning

   AND gives ∅. OR gives (−∞, 0) ∪ [2, ∞).

   No real number can satisfy both separated conditions. Inclusive OR accepts either ray; it still excludes every value from 0 up to but not including 2.

5. Solve |x + 2| ≤ 3 using distance from a named centre.

   Show the answer and reasoning

   −5 ≤ x ≤ 1, or [−5, 1].

   Write the inside expression as x minus −2: the centre is −2. The distance is at most 3, so the endpoints are −2 − 3 = −5 and −2 + 3 = 1, both included.

6. Describe the solutions of |x − 4| < 0, |x − 4| ≤ 0 and |x − 4| > −1.

   Show the answer and reasoning

   Respectively: ∅; only x = 4; every real number.

   Distance is nonnegative. It cannot be less than zero, is at most zero only when it is zero, and is always greater than −1. A positive-radius template cannot be applied indiscriminately.

7. A model requires 4n + 3 ≤ 18, where n is a nonnegative whole-number ticket count. Find every allowed count.

   Show the answer and reasoning

   n = 0, 1, 2 or 3.

   The real inequality gives n ≤ 15/4. The context allows only nonnegative integers. Three gives a model total of 15, while four gives 19 and exceeds the stipulated limit.

What to carry forward

• State the domain and describe every solution, including whether each boundary is included.
• Adding the same defined expression preserves order; multiplying or dividing by a fixed negative number reverses it.
• AND means overlap, while inclusive OR means either condition or both.
• Interpret absolute value as nonnegative distance; check zero and negative thresholds separately.
• Use original-condition checks to detect errors and equivalent reasoning to justify the full solution set.