Solving Absolute Value Inequalities Calculator
This solving absolute value inequalities calculator isolates the bars, then decides between one interval and two rays. For 3|2x + 1| - 4 > 11 the answer is x < -3 or x > 2.
Solution
- Interval notation
- (-∞, -3) ∪ (2, ∞)
- Step 1: isolate the absolute value Add 4 to both sides, then divide both sides by 3.
- 3|2x + 1| - 4 > 11, then 3|2x + 1| > 15, then |2x + 1| > 5
- Step 2: split into two inequalities |u| > 5 means u is more than 5 from zero, in either direction. The answer is two pieces joined by "or".
- 2x + 1 < -5 or 2x + 1 > 5
- Step 3: solve each one Subtract 1, then divide by 2.
- x < -3 or x > 2
- Check The boundary points and a point inside each piece, substituted into the original inequality.
- x = -4: 17 > 11; x = -3: 11 not > 11; x = 2: 11 not > 11; x = 3: 17 > 11
- Graph The solution is the part of the x-axis where the graph is above the line y = 11.
- y = 3|2x + 1| - 4 is a V with its corner at (-1/2, -4)
- Every value is an exact fraction. Decimals are read as written, so 0.1 is exactly 1/10.
How the Absolute Value Calculator works
An absolute value inequalities calculator, or a person with a pencil, solves one in three moves: isolate the absolute value, rewrite it without the bars, and solve what is left. The direction of the inequality decides what "without the bars" looks like.
The default example, 3|2x + 1| - 4 > 11.
1. Isolate. Add 4 to both sides: 3|2x + 1| > 15. Divide by 3: |2x + 1| > 5.
2. Rewrite. |u| > 5 means u is more than 5 from zero, on either side: 2x + 1 < -5 or 2x + 1 > 5.
3. Solve each. 2x < -6 gives x < -3; 2x > 4 gives x > 2.
The answer is x < -3 or x > 2, which is (-∞, -3) ∪ (2, ∞) in interval notation. Check a point from each part: x = -4 gives 3|-7| - 4 = 17, and x = 3 gives 3|7| - 4 = 17, both above 11. A point between, x = 0, gives 3|1| - 4 = -1, which is not.
Less than gives one interval. |u| < R says u is within R of zero, so it becomes one compound inequality, -R < u < R. For |x - 4| ≤ 3 that is -3 ≤ x - 4 ≤ 3, so 1 ≤ x ≤ 7, the closed interval [1, 7].
Greater than gives two rays. |u| > R becomes u < -R or u > R, two pieces joined by "or". Writing it as -R > u > R is a common mistake: no number is both below -R and above R.
The sign rules. Dividing or multiplying both sides by a negative number reverses the inequality, at the isolating step or the solving step. Strict inequalities (< and >) leave the endpoints out, shown with round brackets; ≤ and ≥ include them, shown with square brackets.
Negative and zero right-hand sides. Once isolated, if the right side is negative, |u| < R and |u| ≤ R have no solution, and |u| > R and |u| ≥ R hold for every real number, because an absolute value is never negative. If it is zero, |u| ≤ 0 has the single solution u = 0, and |u| > 0 holds everywhere except there.
Absolute Value Calculator FAQ
How do you solve absolute value inequalities?
Isolate the absolute value. If it is less than R, write -R < u < R. If it is greater than R, write u < -R or u > R. Then solve for x, reversing the sign whenever you divide by a negative number.
What is the answer to |x - 4| ≤ 3?
-3 ≤ x - 4 ≤ 3, so 1 ≤ x ≤ 7, or [1, 7] in interval notation.
When is an absolute value inequality true for all real numbers?
When the isolated absolute value is greater than (or greater than or equal to) a negative number, or greater than or equal to zero. An absolute value is never negative, so the inequality holds for every x.
Why is a greater-than inequality an or statement?
Because the inside can be far from zero on the negative side or on the positive side, and those are two separate regions of the number line. No single interval contains both.