H Algebra 2/Trig · Unit 1
Solve absolute value inequalities, write the answer in interval notation, and graph it on a number line.
Less than is AND
|A| < k is −k < A < k. Graph a segment. Write an interval.
Greater than is OR
|A| > k is A < −k or A > k. Graph two rays. Write a union of intervals.
Isolate first
Move extra terms, then divide. Reverse the inequality if you divide by a negative.
Open or closed
< or > uses an open circle and a round interval. ≤ or ≥ uses a closed circle and a square bracket.
Already isolated. Integer endpoints. Graph and write interval notation.
The formula is
|A| < k means −k < A < k (use ≤ when the inequality is closed).
|x − 2| < 3
−3 < x − 2 < 3
Add 2 to all three parts.
−1 < x < 5
(−1, 5)
Answer: (−1, 5)
The formula is
|A| < k means −k < A < k (use ≤ when the inequality is closed).
|x + 1| < 4
−4 < x + 1 < 4
Subtract 1 from all three parts.
−5 < x < 3
(−5, 3)
Answer: (−5, 3)
The formula is
|A| < k means −k < A < k (use ≤ when the inequality is closed).
|x − 5| ≤ 2
−2 ≤ x − 5 ≤ 2
Add 5 to all three parts.
3 ≤ x ≤ 7
[3, 7]
Answer: [3, 7]
The formula is
|A| > k means A < −k or A > k (use ≤/≥ when closed).
|x + 3| > 2
x + 3 < −2 or x + 3 > 2
Subtract 3 from both sides of each.
x < −5 or x > −1
(−∞, −5) ∪ (−1, ∞)
Answer: (−∞, −5) ∪ (−1, ∞)
The formula is
|A| > k means A < −k or A > k (use ≤/≥ when closed).
|x − 1| ≥ 4
x − 1 ≤ −4 or x − 1 ≥ 4
Add 1 to both sides of each.
x ≤ −3 or x ≥ 5
(−∞, −3] ∪ [5, ∞)
Answer: (−∞, −3] ∪ [5, ∞)
The formula is
|A| < k means −k < A < k (use ≤ when the inequality is closed).
|x + 4| ≤ 1
−1 ≤ x + 4 ≤ 1
Subtract 4 from all three parts.
−5 ≤ x ≤ −3
[−5, −3]
Answer: [−5, −3]
Isolate, then split. Endpoints may be fractions.
Isolate the absolute value.
1|x + 1| < 6
The formula is
|A| < k means −k < A < k (use ≤ when the inequality is closed).
|x + 1| < 6
−6 < x + 1 < 6
Subtract 1 from all three parts.
−7 < x < 5
(−7, 5)
Answer: (−7, 5)
Isolate the absolute value.
|x − 3| ≤ 4
The formula is
|A| < k means −k < A < k (use ≤ when the inequality is closed).
|x − 3| ≤ 4
−4 ≤ x − 3 ≤ 4
Add 3 to all three parts.
−1 ≤ x ≤ 7
[−1, 7]
Answer: [−1, 7]
Isolate the absolute value.
1|2x + 1| ≥ 6
The formula is
|A| > k means A < −k or A > k (use ≤/≥ when closed).
|2x + 1| ≥ 6
2x + 1 ≤ −6 or 2x + 1 ≥ 6
Subtract 1 from both sides of each.
Divide by 2.
x ≤ −72 or x ≥ 52
(−∞, −72] ∪ [52, ∞)
Answer: (−∞, −72] ∪ [52, ∞)
Isolate the absolute value.
3|x + 2| < 9
|x + 2| < 3
The formula is
|A| < k means −k < A < k (use ≤ when the inequality is closed).
|x + 2| < 3
−3 < x + 2 < 3
Subtract 2 from all three parts.
−5 < x < 1
(−5, 1)
Answer: (−5, 1)
Isolate the absolute value.
4|x − 1| ≤ 12
|x − 1| ≤ 3
The formula is
|A| < k means −k < A < k (use ≤ when the inequality is closed).
|x − 1| ≤ 3
−3 ≤ x − 1 ≤ 3
Add 1 to all three parts.
−2 ≤ x ≤ 4
[−2, 4]
Answer: [−2, 4]
The formula is
|A| > k means A < −k or A > k (use ≤/≥ when closed).
|3x − 2| ≥ 4
3x − 2 ≤ −4 or 3x − 2 ≥ 4
Add 2 to both sides of each.
Divide by 3.
x ≤ −23 or x ≥ 2
(−∞, −23] ∪ [2, ∞)
Answer: (−∞, −23] ∪ [2, ∞)
A leading negative, or an inequality that is always true or never true.
Isolate the absolute value.
−1|x − 2| < −2
Dividing by a negative reverses the inequality.
|x − 2| > 2
The formula is
|A| > k means A < −k or A > k (use ≤/≥ when closed).
|x − 2| > 2
x − 2 < −2 or x − 2 > 2
Add 2 to both sides of each.
x < 0 or x > 4
(−∞, 0) ∪ (4, ∞)
Answer: (−∞, 0) ∪ (4, ∞)
Isolate the absolute value.
−1|x + 1| ≤ −4
Dividing by a negative reverses the inequality.
|x + 1| ≥ 4
The formula is
|A| > k means A < −k or A > k (use ≤/≥ when closed).
|x + 1| ≥ 4
x + 1 ≤ −4 or x + 1 ≥ 4
Subtract 1 from both sides of each.
x ≤ −5 or x ≥ 3
(−∞, −5] ∪ [3, ∞)
Answer: (−∞, −5] ∪ [3, ∞)
Isolate the absolute value.
4|2x − 1| ≥ −4
|2x − 1| ≥ −1
Absolute value is always at least 0, which is greater than a negative number.
All real numbers.
Answer: all real numbers
Isolate the absolute value.
3|x − 4| < −1
|x − 4| < −13
Absolute value is never negative, so it cannot be less than a negative number.
No solution.
Answer: no solution
Isolate the absolute value.
5|x + 2| ≤ 10
|x + 2| ≤ 2
The formula is
|A| < k means −k < A < k (use ≤ when the inequality is closed).
|x + 2| ≤ 2
−2 ≤ x + 2 ≤ 2
Subtract 2 from all three parts.
−4 ≤ x ≤ 0
[−4, 0]
Answer: [−4, 0]
Isolate the absolute value.
−1|2x + 3| > −3
Dividing by a negative reverses the inequality.
|2x + 3| < 3
The formula is
|A| < k means −k < A < k (use ≤ when the inequality is closed).
|2x + 3| < 3
−3 < 2x + 3 < 3
Subtract 3 from all three parts.
Divide by 2.
−3 < x < 0
(−3, 0)
Answer: (−3, 0)
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