Solving Absolute Value Equations on the SAT
An absolute value equation almost always has two answers, because the expression inside the bars can be positive or negative.
Reading the concept isn't enough. The score comes from practicing the real Bluebook-style format and finding the exact mistakes you keep making.
What the SAT tests here
If $|A| = b$, then $A = b$ or $A = -b$. You set up and solve both cases, then check each answer in the original equation to rule out any that don't actually work.
How to solve one, step by step
Example: solve $|2x - 1| = 7$.
- Case 1: $2x - 1 = 7 \Rightarrow x = 4$.
- Case 2: $2x - 1 = -7 \Rightarrow x = -3$.
- Both check out, so the solutions are $x = 4$ and $x = -3$.
The mistakes that cost points
- Solving only one case. Forgetting the negative case drops half the answer.
- Skipping the check. Some setups produce extraneous solutions that don't satisfy the original equation.
- Distributing the bars. $|2x - 1|$ is not $2|x| - 1$. The absolute value applies to the whole expression.
Practice questions
Try these the way you would on test day, then open the solution to check your method.
Easy
The equation $|2x - k| = 4$ has $x = 1$ as one of its solutions. What is one possible value of $k$?
- A$k = 2$
- B$k = -2$
- C$k = 6$
- D$k = -6$
Show solution
Answer: B, $k = -2$. Substitute $x = 1$: $|2(1) - k| = 4 \Rightarrow |2 - k| = 4$. Two cases: Case 1: $2 - k = 4 \Rightarrow k = -2$. Case 2: $2 - k = -4 \Rightarrow k = 6$. Both are valid values of $k$. A correct answer is $k = -2$.
Medium
What are all solutions to $|3x - 6| = 21$?
- A$x = 9$ or $x = 5$
- B$x = -5$ only
- C$x = 9$ only
- D$x = 9$ or $x = -5$
Show solution
Answer: D, $x = 9$ or $x = -5$. Set up two cases. Case 1: $3x - 6 = 21 \Rightarrow 3x = 27 \Rightarrow x = 9$. Case 2: $3x - 6 = -21 \Rightarrow 3x = -15 \Rightarrow x = -5$. The solutions are $x = 9$ or $x = -5$.
Hard
The equation $|2x - k| = x + 1$ has $x = 4$ as one of its solutions. What is one possible value of $k$?
- A$k = 13$
- B$k = 3$
- C$k = 8$
- D$k = -3$
Show solution
Answer: B, $k = 3$. Substitute $x = 4$ on the right side: $|2(4) - k| = 4 + 1 = 5 \Rightarrow |8 - k| = 5$. Case 1: $8 - k = 5 \Rightarrow k = 3$. Check: $|2(4) - 3| = |5| = 5$ and $x + 1 = 5$ ✓. Case 2: $8 - k = -5 \Rightarrow k = 13$. Check: $|2(4) - 13| = |-5| = 5$ and $x + 1 = 5$ ✓. Both are valid. $k = 3$ is one correct answer.
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