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|x-1|+|x-2|=|x-3| Can you show me the solution to this equation?

user27301
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  • You need to consider cases if you want an algebraic approach. Less formal: Have you graphed $|x-1|+|x-2|-|x-3|=0$? Might help to get an idea. – imranfat Mar 19 '16 at 02:32

1 Answers1

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You would need to consider several cases.

Case 1: $x\geq 3$

Then the equality would simplify as $$ (x-1)+(x-2) = (x-3) \iff 2x = x \iff x=0. $$ Hence, no $x\geq 3$ satisfies the equality.

Case 2: $3>x\geq 2$ would yield $$ (x-1)+(x-2) = 3-x \iff 3x = 6 \iff x = 2. $$ Thus, $x=2$ is the only value that satisfies the equality in this range.

Case 3: $2>x\geq 1$ would yield $$ (x-1)+(2-x) = 3-x \iff x = 2. $$ Thus, nothing works in this range.

Finally,

Case 4: $1>x$ yields $$ (1-x)+(2-x) = 3-x \iff 2x = x \iff x=0, $$ hence, $x=0$ is in this range, thus satisfies the equality.

In short, $x\in \{0,2\}$ satisfy the equality.

Frank
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