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I'm having trouble understanding how the following is solved.

$$|x+1| = |x-2|$$

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  • $\begingroup$ Recall that $|a| = a$ or $-a$, according to the signs. There are four options here. $\endgroup$
    – user61527
    Jun 11 '14 at 0:04
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The other answers provide a good idea about how to deal with absolute value expressions in general, but here there is a more intuitive approach.

$|x-a|$ is just the distance between $x$ and $a$, so when you have $|x-2| = |x+1| = |x-(-1)|$, your solution is a point that's the same distance from $2$ as it is from $-1$.

The only such point is that halfway in between them: $x = \frac{1}{2}(2+(-1)) = \frac{1}{2}$

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You can begin with $$\pm(x+1)=\pm(x-2)$$ which results in 4 different equations:

$$x+1=x-2$$ $$-(x+1)=x-2$$ $$-(x+1)=-(x-2)$$ $$x+1=-(x-2)$$

The first and third have no solutions, but the other two can be solved to get your actual solutions.

In addition, as the comment added states, it would be wise to check solutions in the original equation, just in case.

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  • $\begingroup$ Also, once you get your solution(s), it would be wise to check them in the original equation to make sure that they are valid. $\endgroup$
    – Théophile
    Jun 11 '14 at 0:09
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So we know what absolute value means. It means that we want the positive version of $x+1$ and $x-2$.

So we don't know what the signs of $x-2$ and $x+2$ are.

Thus, we can make four different equalities, of which $0$, $1$, $2$, $3$, or $4$ may be true. There may be multiple values of $x$ or there might not be any value of $x$ that satisfies.

First, $x+1 = x-2$ simply implies that $x+3 = x$ which is not possible.

Secondly, $-(x+1) = x-2$ implies that $2x=1$, so $x = 1/2$.

Thirdly, $(x+1)=-(x-2)$. This case is the same as the second case because we can multiply both sides by $-1$.

Fourthly, $-(x+1)=-(x-2)$ implies that $x+1=x-2$, which is the same as the first case.

Thus, the only possible answer is in cases 2 and 3 for which $x = 1/2$. To check, note that $1+1/2 = 3/2$ and $1/2-2= -3/2$. $3/2$ and $-3/2$ have the same absolute value.

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