SAT Absolute Value Questions
SAT Absolute Value Questions
Equality • Inequality • Digital SAT Shortcuts | SATMath800 — Premium Chapter Draft
Part 1 — What Absolute Value Really Means
Absolute value represents distance from zero on the number line. Distance is never negative.
- |5| = 5
- |−5| = 5
- |0| = 0
SAT Idea: Absolute value tells us how far a number is from zero.
Visualizing Distance
The equation |x − 2| = 5 asks: “Which numbers are 5 units away from 2?”
-3 2 7
●───────────●───────────●
5 units 5 units
So the solutions are x = −3 and x = 7.
Quick Check — Before We Start
| Question | Your Answer |
|---|---|
| |2| = |−2| | |
| |−5| < |4| | |
| If |x| < 0, then x has a real value. | |
| |2| − |−10| = ? | |
| |7 − 9 + 15| − 3 = ? | |
| |−12| + |5| = ? |
Part 2 — SAT Absolute Value Equality Questions
Rule: If |A| = k and k > 0, then A = k or A = −k.
Example 1
Solve |x| = 7.
x = 7 or x = −7
Example 2 — The Two-Branch Method
Solve |2x − 3| = 11.
Branch 1
2x − 3 = 11
2x = 14
x = 7
Check: |2(7) − 3| = |11| = 11
Branch 2
2x − 3 = −11
2x = −8
x = −4
Check: |2(−4) − 3| = |−11| = 11
Final answer: x = 7 or x = −4
Which of the following is the solution for x if |x + 5| = −1?
Absolute value represents distance, and distance can never be negative.
|x + 5| is always greater than or equal to 0. It can never equal −1.
Correct answer: No solution
Example 3 — Absolute Value Mixed with a Linear Expression
Solve |5x + 9| = 3x + 5.
Branch 1 (5x + 9 ≥ 0)
5x + 9 = 3x + 5
Subtract 3x: 2x + 9 = 5
Subtract 9: 2x = −4
Divide by 2: x = −2
Branch 2 (5x + 9 < 0)
−(5x + 9) = 3x + 5
Distribute: −5x − 9 = 3x + 5
Add 5x: −9 = 8x + 5
Subtract 5: −14 = 8x
Divide by 8: x = −7/4
Final answer: No real solution
Part 3 — SAT Absolute Value Inequality Questions
Pattern A — Less Than → AND
Find the sum of all integer solutions of |x + 3| < 4.
Many students first try to think about what it means for x + 3 to be between −4 and 4.
Possible values of x + 3 include:
- x + 3 = −3 → x = −6
- x + 3 = −2 → x = −5
- x + 3 = −1 → x = −4
- x + 3 = 0 → x = −3
- x + 3 = 1 → x = −2
- x + 3 = 2 → x = −1
- x + 3 = 3 → x = 0
This helps us SEE the interval, but it is too slow for the SAT.
Less than means INSIDE the interval, so use AND.
Ideal SAT Solution:
−4 < x + 3 < 4
Subtract 3 from ALL THREE parts:
−4 − 3 < x + 3 − 3 < 4 − 3
−7 < x < 1
So the integer solutions are: {−6, −5, −4, −3, −2, −1, 0}
Add them carefully: −6 + (−5) + (−4) + (−3) + (−2) + (−1) + 0 = −21
Final answer: −21
Pattern B — Greater Than → OR
Solve |x + 1| > 4.
Branch 1
x + 1 > 4
x > 3
Branch 2
x + 1 < −4
x < −5
Final answer: x > 3 or x < −5
Solve |3x| < −10.
|3x| is always greater than or equal to 0.
A nonnegative number can never be less than −10.
Solution set: ∅
Part 4 — Digital SAT–Style Practice & Worked Solutions
These problems are written in a realistic SAT style.
Question 1: |3x − 6| ≤ 12
Answer: −2 ≤ x ≤ 6
Detailed Solution:
Start with the compound inequality: −2 ≤ 3x − 6 ≤ 12 (Wait, −12 ≤ 3x − 6 ≤ 12)
Add 6 to all three parts: −6 ≤ 3x ≤ 18
Divide all three parts by 3: −2 ≤ x ≤ 6
Question 2: |x + 4| = 9
Answer: x = 5 or x = −13
Detailed Solution:
Split into two branches: x + 4 = 9 or x + 4 = −9
Branch 1: x = 5
Branch 2: x = −13
Final answer: x = 5 or x = −13
Question 3: |2x − 1| > 7
Answer: x > 4 or x < −3
Detailed Solution:
Use the OR pattern: 2x − 1 > 7 or 2x − 1 < −7
Left branch: 2x > 8 → x > 4
Right branch: 2x < -6 → x < −3
Final answer: x > 4 or x < −3
Question 4: |4x + 1| = 13
Answer: x = 3 or x = −7/2
Detailed Solution:
Split into two branches: 4x + 1 = 13 or 4x + 1 = −13
Branch 1: 4x = 12 → x = 3
Branch 2: 4x = −14 → x = −14/4 = −7/2
Final answer: x = 3 or x = −7/2
Question 5: |5 − 2x| < 9
Answer: −2 < x < 7
Detailed Solution:
Use the AND pattern: −9 < 5 − 2x < 9
Subtract 5 from all three parts: −14 < −2x < 4
Divide by −2 and reverse both inequality signs: 7 > x > −2
Rewrite in increasing order: −2 < x < 7
Part 5 — Common SAT Mistakes
- Forgetting the negative branch in an equality question
- Using OR instead of AND for |A| < k
- Using AND instead of OR for |A| > k
- Assuming an absolute value expression can be negative
- Failing to check candidate solutions in the original equation
Part 6 — The 15-Second SAT Absolute Value Cheat Sheet
| Type | Pattern |
|---|---|
| |A| = k | A = k or A = −k |
| |A| < k | −k < A < k |
| |A| ≤ k | −k ≤ A ≤ k |
| |A| > k | A > k or A < −k |
| |A| ≥ k | A ≥ k or A ≤ −k |
INSIDE the interval → AND | OUTSIDE the interval → OR
Final SAT Strategy
Whenever you see an absolute value question on the Digital SAT, ask yourself:
- Is this an equality or an inequality?
- If it is an inequality, am I looking for values INSIDE the interval or OUTSIDE the interval?
- Should the final answer use AND or OR?
- Can the right side ever be negative?
Master these four questions and most SAT absolute value problems become pattern-recognition questions rather than difficult algebra problems.
Digital SAT Absolute Value: Comprehensive Practice Problem Sets
Bluebook-Style Exam Questions with Fully Verified Answers and Step-by-Step Solutions
Category 1 — SAT Absolute Value Basics: Evaluate and Compare
This category focuses on fundamental Digital SAT absolute value skills: evaluating absolute values, simplifying expressions, comparing values, and recognizing absolute value as distance.
Question 1: What is the value of |-14| + |6|?
Question 2: What is the value of |9| – |-13|?
Question 3: Which of the following statements is true?
Question 4: What is the value of |12 – 18 + 7| + |-4|?
Question 5: A number x is 7 units from zero on the number line. Which of the following could be the value(s) of x?
Category 1 Answer Key & Detailed Solutions
| Question | Solution Steps & Explanation | Answer |
|---|---|---|
| Q1 | |-14| = 14, |6| = 6. Adding them gives 14 + 6 = 20. | C |
| Q2 | |9| = 9, |-13| = 13. Subtracting gives 9 – 13 = -4. | A |
| Q3 | |-8| = 8 and |5| = 5. Since 8 > 5, the statement |-8| > |5| is true. | C |
| Q4 | Simplify inside the first absolute value: 12 – 18 + 7 = 1, so |1| = 1. Then |-4| = 4. Adding them gives 1 + 4 = 5. | B |
| Q5 | A distance of 7 units from zero means the number can be 7 units to the right (7) or 7 units to the left (-7). | C |
Category 2 — SAT Absolute Value Equality Questions
This category focuses on solving equations of the form |A| = k, recognizing two possible branches, and verifying solutions.
Question 1: What is the solution set of |x – 4| = 9?
Question 2: What is the solution set of |2x + 3| = 11?
Question 3: What is the solution set of |3x – 5| = 16?
Question 4: If |4x + 1| = 15, what is the sum of all real solutions?
Question 5: How many real solutions does the equation |5x – 2| = 18 have?
Category 2 Answer Key & Detailed Solutions
| Question | Solution Steps & Explanation | Answer |
|---|---|---|
| Q1 | Set up two branches: x – 4 = 9 implies x = 13 or x – 4 = -9 implies x = -5. Solution set is {-5, 13}. | C |
| Q2 | Set up two branches: 2x + 3 = 11 implies 2x = 8 implies x = 4 or 2x + 3 = -11 implies 2x = -14 implies x = -7. Solution set is {-7, 4}. | C |
| Q3 | Set up two branches: 3x – 5 = 16 implies 3x = 21 implies x = 7 or 3x – 5 = -16 implies 3x = -11 implies x = -11/3. Solution set is {7, -11/3}. | C |
| Q4 | Branches give 4x + 1 = 15 implies 4x = 14 implies x = 7/2, and 4x + 1 = -15 implies 4x = -16 implies x = -4. Sum: 7/2 – 8/2 = -1/2. | B |
| Q5 | Branches yield 5x – 2 = 18 implies x = 4 and 5x – 2 = -18 implies x = -16/5, yielding 2 distinct real solutions. | C |
Category 3 — SAT Absolute Value Impossible Cases
This category focuses on recognizing when an absolute value equation or inequality has no real solution because absolute value outputs are never negative.
Question 1: What is the solution set of |x + 7| = -4?
Question 2: How many real solutions does |2x – 5| = -9 have?
Question 3: What is the solution set of |3x + 1| < -2?
Question 4: Which statement is always true for any real number expression A?
Question 5: What is the solution set of |4x – 8| = -1?
Category 3 Answer Key & Detailed Solutions
| Question | Solution Steps & Explanation | Answer |
|---|---|---|
| Q1 | Absolute value represents distance and can never equal a negative number like -4. | D |
| Q2 | Absolute value is always ≥ 0, so it can never equal -9, resulting in 0 solutions. | A |
| Q3 | A nonnegative value (|3x + 1| ≥ 0) can never be less than -2, yielding an empty set. | D |
| Q4 | By definition, absolute value outputs are never negative and include zero (|0| = 0), so |A| ≥ 0. | C |
| Q5 | The left side must be ≥ 0, so it cannot equal -1, yielding no solution (∅). | D |
Category 4 — SAT Absolute Value Equations with Linear Expressions
Solving equations where an absolute value equals a linear expression, verifying whether algebraic branches produce valid solutions.
Question 1: What is the solution set of |x + 3| = x + 7?
Question 2: What is the solution set of |2x – 1| = x + 5?
Question 3: How many real solutions does |3x + 4| = 2x – 1 have?
Question 4: What is the solution set of |x – 5| = 2x + 1?
Question 5: What is the solution set of |4x + 2| = x + 8?
Category 4 Answer Key & Detailed Solutions
| Question | Solution Steps & Explanation | Answer |
|---|---|---|
| Q1 | Branch 1 (x + 3 = x + 7 implies 3 = 7) has no solution. Branch 2: -(x + 3) = x + 7 implies -x – 3 = x + 7 implies 2x = -10 implies x = -5. Verifying: |-5 + 3| = |-2| = 2, and -5 + 7 = 2 (Valid). | A |
| Q2 | Branch 1: 2x – 1 = x + 5 implies x = 6 (Check: |12-1| = 11, 6+5=11, valid). Branch 2: -(2x – 1) = x + 5 implies -2x + 1 = x + 5 implies 3x = -4 implies x = -4/3 (Check: |-8/3 – 1| = |-11/3| = 11/3, -4/3 + 5 = 11/3, valid). | C |
| Q3 | The right side requires 2x – 1 ≥ 0 implies x ≥ 1/2. Branch 1: 3x + 4 = 2x – 1 implies x = -5 (extraneous since -5 < 1/2). Branch 2: 3x + 4 = -(2x – 1) implies 5x = -3 implies x = -3/5 (extraneous). Neither solution satisfies the condition, yielding 0 solutions. | A |
| Q4 | Right side requires 2x + 1 ≥ 0 implies x ≥ -1/2. Branch 1: x – 5 = 2x + 1 implies x = -6 (invalid since -6 < -1/2). Branch 2: -(x – 5) = 2x + 1 implies -x + 5 = 2x + 1 implies 3x = 4 implies x = 4/3 (Valid, 4/3 ≥ -1/2). | A |
| Q5 | Branch 1: 4x + 2 = x + 8 implies 3x = 6 implies x = 2 (Valid: |10| = 10). Branch 2: 4x + 2 = -(x + 8) implies 4x + 2 = -x – 8 implies 5x = -10 implies x = -2 (Valid: |-6| = 6). Both x = 2 and x = -2 are valid. | C |
Category 5 — SAT Absolute Value Less-Than Inequalities
Solving |A| < k and |A| ≤ k questions, understanding that the solution stays inside an interval (AND pattern).
Question 1: What is the solution set of |x – 2| < 5?
Question 2: What is the solution set of |2x + 1| ≤ 7?
Question 3: Which of the following lists all integer solutions of |x + 4| < 6?
Question 4: What is the sum of all integer solutions of |x – 1| < 4?
Question 5: What is the solution set of |3x – 6| < 12?
Category 5 Answer Key & Detailed Solutions
| Question | Solution Steps & Explanation | Answer |
|---|---|---|
| Q1 | Set up compound inequality: -5 < x – 2 < 5 implies -3 < x < 7. | B |
| Q2 | Set up compound inequality: -7 ≤ 2x + 1 ≤ 7 implies -8 ≤ 2x ≤ 6 implies -4 ≤ x ≤ 3. | A |
| Q3 | Inequality -6 < x + 4 < 6 implies -10 < x < 2, giving integers from -9 to 1 inclusive. | B |
| Q4 | Interval -4 < x – 1 < 4 implies -3 < x < 5, yielding integers {-2, -1, 0, 1, 2, 3, 4}, which sum to 7. | C |
| Q5 | Setup -12 < 3x – 6 < 12 implies -6 < 3x < 18 implies -2 < x < 6. | A |
Category 6 — SAT Absolute Value Greater-Than Inequalities
Solving |A| > k and |A| ≥ k inequalities, understanding that solutions lie outside an interval (OR pattern).
Question 1: What is the solution set of |x – 1| > 4?
Question 2: What is the solution set of |2x + 3| ≥ 7?
Question 3: What is the solution set of |3x – 6| > 12?
Question 4: Which of the following lists all integer solutions of |x + 2| ≥ 5?
Question 5: How many integer solutions satisfy |2x – 5| > 7 for -5 ≤ x ≤ 8?
Category 6 Answer Key & Detailed Solutions
| Question | Solution Steps & Explanation | Answer |
|---|---|---|
| Q1 | Set up branches: x – 1 > 4 implies x > 5 or x – 1 < -4 implies x < -3. | B |
| Q2 | Set up branches: 2x + 3 ≥ 7 implies 2x ≥ 4 implies x ≥ 2 or 2x + 3 ≤ -7 implies 2x ≤ -10 implies x ≤ -5. | B |
| Q3 | Set up branches: 3x – 6 > 12 implies 3x > 18 implies x > 6 or 3x – 6 < -12 implies 3x < -6 implies x < -2. | B |
| Q4 | Set up branches: x + 2 ≥ 5 implies x ≥ 3 or x + 2 ≤ -5 implies x ≤ -7. Integers are x ≤ -7 and x ≥ 3. | B |
| Q5 | Branches give 2x – 5 > 7 implies 2x > 12 implies x > 6 (integers in range: 7, 8), and 2x – 5 < -7 implies 2x < -2 implies x < -1 (integers in range: -5, -4, -3, -2). Total restricted integers: {-5, -4, -3, -2} and {7, 8}, totaling 6 integers. | B |
Category 7 — SAT Absolute Value Mixed Challenge Problems
Combining multiple absolute value skills: equality, inequality, impossible cases, and equations requiring candidate verification.
Question 1: What is the sum of all real solutions of |2x – 7| = 9?
Question 2: How many real solutions does |4x + 1| = -5 have?
Question 3: What is the solution set of |x – 4| ≤ 3?
Question 4: How many integer solutions satisfy |x + 1| < 5?
Question 5: What is the solution set of |x – 2| = x + 4?
Category 7 Answer Key & Detailed Solutions
| Question | Solution Steps & Explanation | Answer |
|---|---|---|
| Q1 | Branches yield 2x – 7 = 9 implies 2x = 16 implies x = 8, and 2x – 7 = -9 implies 2x = -2 implies x = -1. Sum equals 8 + (-1) = 7. | A |
| Q2 | Absolute value can never equal a negative number like -5, resulting in 0 solutions. | A |
| Q3 | Compound interval -3 ≤ x – 4 ≤ 3 implies 1 ≤ x ≤ 7. | A |
| Q4 | Interval -5 < x + 1 < 5 implies -6 < x < 4, giving 9 integer solutions ({-5, -4, -3, -2, -1, 0, 1, 2, 3}). | C |
| Q5 | Right side requires x + 4 ≥ 0 implies x ≥ -4. Branch 1: x – 2 = x + 4 implies -2 = 4 (no solution). Branch 2: -(x – 2) = x + 4 implies -x + 2 = x + 4 implies 2x = -2 implies x = -1 (Valid, since -1 ≥ -4). | A |
