AMC 10 · 2004 · #4
Grade 7 algebraWhat is the value of x if ∣x−1∣=∣x−2∣?
Pick an answer.
AMC 10 2004 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 the number $x$ whose distance to $1$ equals its distance to $2$, written as the equation $|x-1|=|x-2|$.
Givens: The equation $|x-1|=|x-2|$ must hold.; $x$ is a real number.; Answer choices: (A) $-\frac12$, (B) $\frac12$, (C) $1$, (D) $\frac32$, (E) $2$
Unknowns: The value of $x$ that makes the two absolute values equal.
Understand
Restated: Find the number $x$ whose distance to $1$ equals its distance to $2$, written as the equation $|x-1|=|x-2|$.
Givens: The equation $|x-1|=|x-2|$ must hold.; $x$ is a real number.; Answer choices: (A) $-\frac12$, (B) $\frac12$, (C) $1$, (D) $\frac32$, (E) $2$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #6 Guess and Check, #4 Introduce a Variable
The two absolute values are begging to be read as distances on a number line (Tool #1): $|x-1|$ is how far $x$ sits from $1$, and $|x-2|$ is how far it sits from $2$. Wanting those distances equal turns the algebra into a one-line geometry fact — the point equidistant from $1$ and $2$ is the point halfway between them. Tool #6 (Guess and Check) then confirms the candidate by plugging it back in, and Tool #4 (Introduce a Variable) offers a purely algebraic cross-check by squaring both sides.
Execute — Answer: D
7.NS.A.1 Step 1 Read each side as a distance
- On the number line, $|x-1|$ measures the distance from $x$ to $1$, and $|x-2|$ measures the distance from $x$ to $2$.
- The equation $|x-1|=|x-2|$ therefore says: $x$ is the same distance from $1$ as it is from $2$.
- Picture $1$ and $2$ marked on the line and $x$ somewhere that keeps both gaps equal.
💡 The distance between two numbers on the line is the absolute value of their difference.
5.NF.B.3 Step 2 The equidistant point is the midpoint
- A point that is equally far from $1$ and from $2$ must sit exactly halfway between them.
- The halfway point of two numbers is their average, so take $x = \frac{1+2}{2}$.
- Adding gives $3$, and dividing by $2$ gives $\frac32$.
- So $x=\frac32$.
💡 Halfway between two spots on a line is just the average of the two numbers.
6.NS.C.7 Step 3 Check it and name the choice
- Test $x=\frac32$ in the original equation.
- The left side is $\left|\frac32-1\right| = \left|\frac12\right| = \frac12$, and the right side is $\left|\frac32-2\right| = \left|-\frac12\right| = \frac12$.
- The two sides match, so $\frac32$ really is equidistant from $1$ and $2$.
- That value is choice (D).
💡 Plugging the candidate back in and getting equal lengths on both sides confirms the answer.
7.NS.A.1 On the number line, $|x-1|$ measures the distance from $x$ to $1$, and $|x-2|$ m 5.NF.B.3 A point that is equally far from $1$ and from $2$ must sit exactly halfway betwe 6.NS.C.7 Test $x=\frac32$ in the original equation. The left side is $\left|\frac32-1\rig Review
Reasonableness: The solution must lie strictly between $1$ and $2$, since only a point inside that gap can be equally far from both endpoints. That instantly rules out (A) $-\frac12$, (B) $\frac12$, (C) $1$, and (E) $2$, all of which sit on or outside the ends. Only (D) $\frac32$ lands in the middle, and $\frac32=1.5$ is exactly halfway, so the answer is reasonable.
Alternative: Solve it with pure algebra. Both sides of $|x-1|=|x-2|$ are non-negative, so squaring is safe: $(x-1)^2=(x-2)^2$ gives $x^2-2x+1 = x^2-4x+4$. The $x^2$ terms cancel, leaving $-2x+1 = -4x+4$, so $2x = 3$ and $x=\frac32$. Reaching the same $\frac32$ by a different route confirms (D).
CCSS standards used (min grade 7)
7.NS.A.1Apply and extend understanding of addition and subtraction to rational numbers (Reading $|x-1|$ and $|x-2|$ as the distances from $x$ to $1$ and to $2$ — the distance between two numbers is the absolute value of their difference.)5.NF.B.3Interpret a fraction as division of the numerator by the denominator (Finding the midpoint as the average $\frac{1+2}{2}=\frac32$.)6.NS.C.7Understand ordering and absolute value of rational numbers (Evaluating $\left|\frac12\right|$ and $\left|-\frac12\right|$ to check that both sides equal $\frac12$.)
⭐ When a point is the same distance from two numbers on the line, it sits exactly halfway between them — the average of the two.
⭐ When a point is the same distance from two numbers on the line, it sits exactly halfway between them — the average of the two.
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