Solving Absolute Value Inequalities

Absolute value is distance, so inequalities become questions about which numbers lie near or far from a center. Remembering that picture helps prevent the common mistake of using the same compound form for less-than and greater-than problems.

Less than means inside the interval

For |x − 3| < 5, the distance between x and 3 is less than 5. Write −5 < x − 3 < 5, then add 3 to every part: −2 < x < 8. The solution is the interval between the two endpoints, so both inequalities must be true (AND).

Greater than means outside the interval

For |x − 3| > 5, x is more than 5 units away from 3. Split it into x − 3 < −5 or x − 3 > 5, then solve: x < −2 or x > 8. Either outside region works (OR).

Handle nonpositive right sides

Absolute value is never negative. Therefore |expression| < a negative number has no solution, while |expression| > a negative number is true for every real value, subject to any domain restrictions. For ≤ 0, solve where the expression inside equals zero; for ≥ 0, every value satisfies the inequality.

Graph and verify

• Draw the center and endpoints on a number line.
• Use a closed endpoint for ≤ or ≥ and an open endpoint for < or >.
• Shade between endpoints for a less-than distance condition.
• Shade outside endpoints for a greater-than distance condition.
• Test a value from the shaded region in the original inequality.

Solve with a coefficient and a shifted center

For |2x − 1| ≤ 7, first recognize the center by rewriting the expression as 2(x − 1/2). Since the right side is nonnegative, use −7 ≤ 2x − 1 ≤ 7. Add 1 to every part to get −6 ≤ 2x ≤ 8, then divide by positive 2: −3 ≤ x ≤ 4. The interval is closed because the original comparison includes equality.

Watch the sign when dividing

After converting the absolute value inequality to a compound inequality, ordinary inequality rules still apply. If a step divides by a negative coefficient, reverse both inequality signs. For |−2x + 1| < 5, write −5 < −2x + 1 < 5, subtract 1, and divide by −2 with both signs reversed: −2 < x < 3. Testing a value such as x = 0 confirms that 1 is less than 5.

Handle equality and impossible bounds

The symbol determines whether endpoints belong to the solution. |x − 2| ≤ 3 includes x = −1 and x = 5; |x − 2| < 3 excludes them. If the right side is negative, an absolute value cannot be less than or equal to it, so those comparisons have no solution. It is always greater than a negative value, so a strict greater-than comparison is true for all inputs in the expression's domain.

Interpret absolute value as distance

The expression |x − a| measures distance between x and a on the number line. “Within r units” means between a−r and a+r. “More than r units away” means outside that interval on either side. The words AND and OR follow the geometry: a point inside must satisfy both bounds; a point outside can be left of the lower bound or right of the upper bound.

Graph and verify

Draw the center and endpoints, choose open circles for strict inequalities and closed circles when equality is included, then shade the inside or outside region. Substitute a point from each shaded region into the original absolute-value statement. This check catches a reversed interval, a missing endpoint, or an error introduced while dividing by a negative.

Translate absolute value into distance

Absolute value measures distance from zero, so |x − a| represents the distance between x and a. The inequality |x − 4| < 3 asks for values less than 3 units from 4. That means −3 < x − 4 < 3; add 4 throughout to get 1 < x < 7. The solution is the interval between the two boundaries.

For a greater-than inequality, the values are outside the distance interval. |x − 4| > 3 means x − 4 < −3 or x − 4 > 3, so x < 1 or x > 7. The connector changes: “less than” absolute distance produces an inside interval with and; “greater than” produces two outside intervals with or.

Include or exclude endpoints correctly

A strict symbol (< or >) excludes the boundary and uses open endpoints. Inclusive symbols (≤ or ≥) include the boundaries and use closed endpoints. Thus |x − 4| ≤ 3 gives 1 ≤ x ≤ 7, including values exactly three units from 4. For |x − 4| ≥ 3, use x ≤ 1 or x ≥ 7.

A constant absolute value such as |x| < 0 has no solution because absolute value cannot be negative and cannot be less than zero. |x| ≥ 0 is true for every real x. Check the constant and the direction before splitting into cases; a universal rule may simplify the work.

Solve algebraically and graph the result

For |2x + 1| ≤ 5, write −5 ≤ 2x + 1 ≤ 5. Subtract 1 throughout: −6 ≤ 2x ≤ 4. Divide by positive 2: −3 ≤ x ≤ 2. If dividing through by a negative, reverse both inequality symbols. Graph the interval with closed circles at −3 and 2 and shade between.

For an absolute value expression not centered on x, isolate the absolute value first using inverse operations. If 2|x − 1| + 3 < 9, subtract 3 and divide by 2 to get |x − 1| < 3, then solve the distance inequality. Keep each step equivalent and test a value from the final interval in the original inequality.

• Interpret |x − a| as distance from a.
• For < or ≤, solve an inside compound inequality using and.
• For > or ≥, split into two outside cases using or.
• Use endpoint inclusion to choose open or closed circles.
• Isolate the absolute value first and verify a test value.

Exam takeaway

Less-than absolute value inequalities create an AND statement between endpoints; greater-than inequalities create an OR statement outside them. Think distance and test a point to verify the solution.

Translate distance into an interval

The inequality |x − 5| < 2 means x is less than two units from 5. Write −2 < x − 5 < 2, then add 5 throughout: 3 < x < 7. The word “less” creates the interval between the two boundary values. On a number line, use open circles because the endpoints are not included.

A greater-than inequality creates two rays

The statement |x − 5| > 2 means x is more than two units away from 5, so x − 5 < −2 or x − 5 > 2. Therefore x < 3 or x > 7. The solution lies outside the interval, with open endpoints for a strict inequality. This is the opposite structure from an absolute value less than a positive number.

Include endpoints for less-than-or-equal

For |x + 1| ≤ 4, write −4 ≤ x + 1 ≤ 4. Subtract 1 to obtain −5 ≤ x ≤ 3. Both boundary values are included because the inequality allows a distance of exactly four. Brackets in interval notation, [−5, 3], express the same inclusion.

Check the zero and negative cases

Absolute value is always nonnegative. Thus |A| < −2 and |A| ≤ −2 have no solutions. For |A| > −2 or |A| ≥ −2, every real value of A works, because every absolute value is at least zero. Recognizing these cases before splitting can prevent extra algebra and endpoint mistakes.

Solve inequalities with a coefficient outside

For 3|x − 2| + 1 ≥ 10, subtract 1 and divide by 3 to get |x − 2| ≥ 3. Split into x − 2 ≤ −3 or x − 2 ≥ 3, giving x ≤ −1 or x ≥ 5. If dividing by a negative coefficient, reverse the inequality sign as usual; isolate the absolute-value expression carefully before applying the distance cases.

Test one point from each proposed region and one from the excluded middle when practical. For the last example, x = 0 satisfies the original inequality, while x = 2 does not. A number-line sketch helps confirm that “greater than” produced two outside rays rather than the interval between the boundaries.