AMC 10 · 2010 · #13
Grade 8 algebraWhat is the sum of all the solutions of x=∣2x−∣60−2x∣∣?
Pick an answer.
AMC 10 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Find every real number $x$ that satisfies $x = \left|\,2x - |60 - 2x|\,\right|$, then add all of those values of $x$ together.
Givens: The equation $x = \left|\,2x - |60 - 2x|\,\right|$; It contains one absolute value nested inside another; Answer choices: (A) $32$, (B) $60$, (C) $92$, (D) $120$, (E) $124$
Unknowns: All values of $x$ that make the equation true; The sum of those values
Understand
Restated: Find every real number $x$ that satisfies $x = \left|\,2x - |60 - 2x|\,\right|$, then add all of those values of $x$ together.
Givens: The equation $x = \left|\,2x - |60 - 2x|\,\right|$; It contains one absolute value nested inside another; Answer choices: (A) $32$, (B) $60$, (C) $92$, (D) $120$, (E) $124$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #3 Eliminate Possibilities
An absolute value is really two rules in disguise: $|A| = A$ when $A \ge 0$ and $|A| = -A$ when $A < 0$. With one absolute value inside another, Tool #7 (Identify Subproblems) splits the tangle into a few clean linear equations, one per sign region. Tool #3 (Eliminate Possibilities) then throws out any answer that lands outside the region it was built for, so only the genuine solutions survive to be added.
Execute — Answer: C
6.NS.C.7 Step 1 The value must be non-negative
- The whole right side sits inside an outer absolute value, and an absolute value is never negative.
- So any solution must satisfy $x \ge 0$.
- This will let us reject impossible candidates later.
💡 Distance from zero is never negative, so $x$ cannot be negative.
6.NS.C.7 Step 2 Split on the inner absolute value
- Start with the inside piece $|60 - 2x|$.
- It changes rule where $60 - 2x = 0$, that is at $x = 30$.
- When $x \le 30$ the inside is non-negative, so $|60 - 2x| = 60 - 2x$; then $2x - |60 - 2x| = 2x - (60 - 2x) = 4x - 60$, and the equation becomes $x = |4x - 60|$.
💡 Peeling one absolute value at a time turns a scary nested expression into an ordinary one.
8.EE.C.7 Step 3 Solve the first case
- Now split $|4x - 60|$ at $x = 15$.
- If $x \ge 15$ then $|4x - 60| = 4x - 60$, so $x = 4x - 60$ gives $3x = 60$, $x = 20$; since $15 \le 20 \le 30$ this is valid.
- If $x < 15$ then $|4x - 60| = 60 - 4x$, so $x = 60 - 4x$ gives $5x = 60$, $x = 12$; since $12 < 15$ this is valid too.
💡 Each sign choice gives a plain linear equation you can solve in one line.
8.EE.C.7 Step 4 Solve the second case
- When $x > 30$ the inside is negative, so $|60 - 2x| = 2x - 60$.
- Then $2x - |60 - 2x| = 2x - (2x - 60) = 60$, and the equation becomes $x = |60| = 60$.
- Since $60 > 30$, this fits its own range, so $x = 60$ is valid.
💡 In this range the two $2x$ terms cancel, leaving a constant that answers itself.
6.EE.B.5 Step 5 Collect and add the solutions
- Every candidate ($12$, $20$, $60$) is non-negative and lies inside the range it came from, so all three are real solutions.
- Their sum is $12 + 20 + 60 = 92$, which is choice (C).
💡 Keep only the values that satisfy the original equation, then add them for the final answer.
6.NS.C.7 The whole right side sits inside an outer absolute value, and an absolute value 6.NS.C.7 Start with the inside piece $|60 - 2x|$. It changes rule where $60 - 2x = 0$, th 8.EE.C.7 Now split $|4x - 60|$ at $x = 15$. If $x \ge 15$ then $|4x - 60| = 4x - 60$, so 8.EE.C.7 When $x > 30$ the inside is negative, so $|60 - 2x| = 2x - 60$. Then $2x - |60 - 6.EE.B.5 Every candidate ($12$, $20$, $60$) is non-negative and lies inside the range it Review
Reasonableness: Test each solution in the original equation. $x = 12$: $|60 - 24| = 36$, then $|24 - 36| = 12$. Correct. $x = 20$: $|60 - 40| = 20$, then $|40 - 20| = 20$. Correct. $x = 60$: $|60 - 120| = 60$, then $|120 - 60| = 60$. Correct. All three check out, and $12 + 20 + 60 = 92$ matches choice (C).
Alternative: Tool #13 (Convert to Algebra): rewrite $x = |2x - |60 - 2x||$ as $2x - |60 - 2x| = \pm x$, so $|60 - 2x| = 2x \mp x$, giving $|60 - 2x| = x$ or $|60 - 2x| = 3x$. Dropping the last absolute value produces four linear equations $60 - 2x = \pm x$ and $60 - 2x = \pm 3x$; solving and discarding the one negative (extraneous) root again yields $12$, $20$, $60$ with sum $92$.
CCSS standards used (min grade 8)
6.NS.C.7Understand ordering and absolute value of rational numbers (Reading each absolute value as its distance-from-zero rule, noting the right side is $\ge 0$, and rewriting $|60 - 2x|$ correctly on each side of $x = 30$.)8.EE.C.7Solve linear equations in one variable (Solving the linear equations that each sign case produces: $x = 4x - 60$, $x = 60 - 4x$, and $x = 60$.)6.EE.B.5Understand solving an equation or inequality as a process of finding values (Checking that each candidate actually lies in the range it was built for, keeping $12$, $20$, and $60$, then summing them.)
⭐ Peel a nested absolute value one layer at a time: each sign choice becomes a simple line equation, and you keep only the answers that land in their own range.
⭐ Peel a nested absolute value one layer at a time: each sign choice becomes a simple line equation, and you keep only the answers that land in their own range.
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