2004 AMC 10A Problems/Problem 4

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Problem

What is the value of $x$ if $|x-1|=|x-2|$?

$\mathrm{(A) \ } -\frac12 \qquad \mathrm{(B) \ } \frac12 \qquad \mathrm{(C) \ } 1 \qquad \mathrm{(D) \ } \frac32 \qquad \mathrm{(E) \ } 2$

Solution

$|x-1|$ is the distance between $x$ and $1$; $|x-2|$ is the distance between $x$ and $2$.

Therefore, the given equation says $x$ is equidistant from $1$ and $2$, so $x=\frac{1+2}2=\frac{3}{2}\Rightarrow\boxed{\mathrm{(D)}\ \frac{3}{2}}$.

Alternatively, we can solve by casework (a method which should work for any similar problem involving absolute values of real numbers). If $x \leq 1$, then $|x - 1| = 1-x$ and $|x - 2| = 2 - x$, so we must solve $1 - x = 2 - x$, which has no solutions. Similarly, if $x \geq 2$, then $|x - 1| = x - 1$ and $|x - 2| = x - 2$, so we must solve $x - 1 = x- 2$, which also has no solutions. Finally, if $1 \leq x \leq 2$, then $|x - 1| = x - 1$ and $|x - 2| = 2-x$, so we must solve $x - 1 = 2 - x$, which has the unique solution $x = \frac32$.

Solution 2

We know that either $x-1=x-2$ or $x-1=-(x-2)$. The first equation simplifies to $-1=-2$, which is false, so $x-1=-(x-2)$. Solving, we get $x=\frac{3}{2}\Rightarrow\boxed{\mathrm{(D)}\ \frac{3}{2}}$.

See also

2004 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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All AMC 10 Problems and Solutions

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