Absolute Value Inequalities

Absolute value inequalities describe numbers whose distance from a reference point is less than, at most, greater than or at least a specified amount. In brief: |x − 2| < 3 means −1 < x < 5, while |x − 2| > 3 means x < −1 or x > 5. The first selects an inside interval; the second selects two outside regions.

A distance picture helps explain the algebraic rules and the difference between and and or. Before using those rules, isolate the absolute value and inspect the comparison value. This guide covers strict and inclusive endpoints, negative coefficients, zero and negative bounds, and checks with representative numbers.

What does an absolute value inequality mean geometrically?

The expression |x − a| measures the distance from x to a on the number line. Requiring that distance to be less than a positive radius r selects points between a − r and a + r. Requiring it to be greater than r selects points farther away on either side of those boundaries.

For |x − 2| < 3, the centre is 2 and the radius is 3. The boundary points are 2 − 3 = −1 and 2 + 3 = 5. Every point strictly between them is within three units of 2, while the boundary points themselves are excluded because their distance is exactly 3.

Changing the symbol to ≤ includes both boundary points. Changing it to > selects the exterior without the boundaries, while ≥ selects the exterior with the boundaries. The sign of the inequality therefore determines both the selected region and whether equality at its endpoints is permitted.

How do you solve absolute value inequalities with less than?

For a positive number r, |A| < r is equivalent to −r < A < r. Both comparisons must hold at once, so this is an and condition. Similarly, |A| ≤ r becomes −r ≤ A ≤ r, with inclusive endpoints retained throughout the solution process.

Solve |2x − 1| < 5 by writing −5 < 2x − 1 < 5. Add 1 to all three parts to obtain −4 < 2x < 6, then divide all three parts by 2. The solution is −2 < x < 3, or the open interval (−2, 3).

A check at x = 0 gives |−1| = 1 < 5, so an interior point works. At the endpoints x = −2 and x = 3, the absolute value equals 5, which is not allowed by the strict inequality. This confirms both the location and the exclusion of the two boundaries.

How do you solve absolute value inequalities with greater than?

For a positive number r, |A| > r is equivalent to A < −r or A > r. Only one branch needs to hold, so the solution is a union of two regions. For |A| ≥ r, use A ≤ −r or A ≥ r, keeping the equality possibilities.

Solve |3x + 2| ≥ 8. The first branch is 3x + 2 ≤ −8, giving x ≤ −10/3. The second is 3x + 2 ≥ 8, giving x ≥ 2. The complete solution is x ≤ −10/3 or x ≥ 2, including both boundary values.

Do not write −8 ≥ 3x + 2 ≥ 8, which would demand mutually incompatible comparisons. Greater-than absolute value conditions select either a sufficiently negative inner value or a sufficiently positive one. The word or and the two separate inequalities preserve that meaning clearly.

How do outside constants and negative coefficients affect the method?

Isolate the absolute value before applying the inside-or-outside rule. For 2|x − 4| + 1 ≤ 7, subtract 1 and divide by 2 to get |x − 4| ≤ 3. Then −3 ≤ x − 4 ≤ 3 gives 1 ≤ x ≤ 7.

Multiplying or dividing an inequality by a negative number reverses its direction. In −2|x + 1| < −6, dividing by −2 gives |x + 1| > 3. This is now an outside condition, so x + 1 < −3 or x + 1 > 3, giving x < −4 or x > 2.

The reversal is essential because multiplying values by a negative number reverses their order on the number line. A check with x = 0 gives −2 < −6, which is false, while x = 3 gives −8 < −6, which is true. These checks support the exterior solution rather than an interior interval.

What happens when the bound is zero or negative?

Absolute value is always nonnegative, so comparisons with zero have special forms. The inequality |A| < 0 has no solution, while |A| ≤ 0 requires A = 0. The condition |A| ≥ 0 holds wherever A is defined, and |A| > 0 holds wherever A is defined and nonzero.

If the bound is negative, |A| < −2 and |A| ≤ −2 are impossible. In contrast, |A| > −2 and |A| ≥ −2 hold for every input where A is defined. A nonnegative quantity is always greater than a negative quantity, so there is no need to split these cases into interval branches.

Domain restrictions remain in force. If A contains 1/x, a statement such as |1/x| ≥ 0 is true for all real x except zero, not for every real number. The absolute-value operation does not define an inner expression at points where that expression is already undefined.

How do you write solutions in interval notation and on a number line?

An inside solution such as −2 < x < 3 is written (−2, 3) in interval notation. Parentheses exclude endpoints, while square brackets include them: −2 ≤ x ≤ 3 becomes [−2, 3]. Mixed strictness uses one of each, though the basic symmetric absolute-value forms usually give matching endpoint types.

An outside solution such as x < −4 or x > 2 becomes (−∞, −4) ∪ (2, ∞). The union symbol combines the two allowed regions. Infinity is not a real endpoint that can be included, so intervals always use parentheses at −∞ and ∞.

On a number line, open circles show excluded finite endpoints and filled circles show included endpoints. Shade between the circles for an inside condition and outward along two rays for an outside condition. Label the boundary numbers before shading so the diagram records the solved inequality rather than only the original centre and radius.

How do absolute value inequalities describe tolerance?

Suppose a part’s modelled length L must be within 0.2 cm of a target of 15 cm, including the boundary. The condition is |L − 15| ≤ 0.2. Rewriting gives −0.2 ≤ L − 15 ≤ 0.2, so 14.8 ≤ L ≤ 15.2.

A length of 15.1 cm satisfies the condition because its deviation is 0.1 cm. A length of 14.7 cm does not, because its deviation is 0.3 cm. The absolute value treats deviations above and below the target symmetrically, while the chosen comparison symbol specifies whether the limiting deviation is accepted.

This background connects algebraic inequalities with intervals of acceptable measurements, errors and locations. It assumes the tolerance is symmetric around the target. If the allowed increase and decrease differ, a compound inequality such as 14.8 ≤ L ≤ 15.3 may express the requirement more directly than a single absolute-value expression centred at 15.

How do you check absolute value inequality solutions?

Test at least one value inside each proposed region and one outside it when practical. Also substitute the boundary values to see whether equality is allowed. These checks do not replace a valid algebraic derivation, but they are efficient at detecting reversed symbols, missing branches and incorrectly included endpoints.

Keep the distinction between and and or explicit. A less-than distance condition requires the inner value to stay above the negative bound and below the positive bound simultaneously. A greater-than distance condition permits either an inner value below the negative bound or one above the positive bound.

Finally, if the comparison side itself depends on x, do not apply a constant-radius rule without checking its sign. An expression such as |x − 1| < x requires x to be positive before the usual double inequality can hold. Variable bounds may need additional case analysis, while the standard rules here apply directly to fixed nonnegative distances.