Absolute Value Equations on the SAT
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Absolute value is distance from $0$, so it is never negative. An equation like $|x - 3| = 5$ splits into two cases, $x - 3 = 5$ or $x - 3 = -5$, which gives two solutions, $8$ and $-2$.
Before you split, get the absolute value by itself on one side. If it then equals a negative number, there is no solution, and you can move on.
Once students know the two-case rule, these questions go quickly. The SAT also likes to describe $|x - a| \le b$ in words, as "within $b$ of $a$", in questions about tolerances and temperatures.
What to do when you see
- $|x - 3| = 5$. Two cases: $x - 3 = 5$ or $x - 3 = -5$.
- $3|x - 2| + 1 = 13$. Isolate first: $|x - 2| = 4$. Then split.
- $|\ldots| = $ a negative number. No solution.
- $|\ldots| = 0$. Exactly one solution.
- "Within $3$ of $10$" $|x - 10| \le 3$, which means $7 \le x \le 13$.
- A variable on both sides, like $|2x - 6| = x + 3$. Split into two cases, then check both answers in the original equation.
Examples
Isolate, then split
Solve $3|x - 2| + 1 = 13$.
- Subtract $1$ and divide by $3$: $|x - 2| = 4$.
- Split: $x - 2 = 4$ or $x - 2 = -4$.
- $x = 6$ or $x = -2$.
Writing a tolerance
A thermostat keeps a room within $4$ degrees of $68$°F. Write this as an inequality for the temperature $t$.
- "Within $4$ of $68$" means the distance from $t$ to $68$ is at most $4$.
- $|t - 68| \le 4$.
- That is the same as $64 \le t \le 72$.
A variable on both sides
Solve $|2x - 6| = x + 3$.
- Case 1: $2x - 6 = x + 3$, so $x = 9$.
- Case 2: $2x - 6 = -(x + 3)$, so $3x = 3$ and $x = 1$.
- Check both: $|12| = 12$ and $|-4| = 4$. Both work.
Common mistakes
Splitting before isolating. In $3|x - 2| + 1 = 13$, get $|x - 2|$ alone first.
Only solving the positive case. Almost every absolute value equation has two solutions.
Solving an impossible equation. $|x + 4| = -2$ has no solution. Write "no solution" and move on.
Negating only part of the right side. The second case of $|2x - 6| = x + 3$ is $2x - 6 = -x - 3$.
Mixing up "within" and "more than". Within $3$ of $10$ is $|x - 10| \le 3$. More than $3$ away is $|x - 10| > 3$.
Your turn
Isolate the absolute value first in every question.
- A$7$ only
- B$-7$ only
- C$-7$ and $7$
- D$0$ and $7$
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Show solution
- A$4$ only
- B$-5$ and $4$
- C$-4$ and $5$
- D$-4$ and $4$
Show solution
- AZero
- BExactly one
- CExactly two
- DInfinitely many
Show solution
Show solution
- A$|x + 10| \le 3$
- B$|x - 3| \le 10$
- C$|x - 10| \ge 3$
- D$|x - 10| \le 3$
Show solution
Show solution
- A$|t + 68| \le 4$
- B$|t - 68| \le 4$
- C$|t - 4| \le 68$
- D$|t - 68| \ge 4$
Show solution
Show solution
Show solution
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