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The following is the equation:

$|x+1|+|x+2|=3$

How can I solve this problem?

Do I have to reformat it to $|x+1|=3-|x-2|$?

I would like a simple answer that by no means uses set theory. The answer must include a step by step explanation

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  • $\begingroup$ You say not to use set theory, but bear in mind, this equation does not have one or two solutions. There will be a whole interval of solutions. It's best to express these solutions as a set, but we can also express it with an inequality if you prefer. Just a heads up. $\endgroup$ – Theo Bendit Apr 8 at 23:09
  • $\begingroup$ But we can express these also with multiple inequalities @TheoBendit $\endgroup$ – BeastCoder2 Apr 8 at 23:10
  • $\begingroup$ Yes, that's true (and I did mention this in my comment). I was trying to confirm that you were just objecting to expressing the solution as a set, not trying to deny the fact that there is an infinite set of solutions. $\endgroup$ – Theo Bendit Apr 8 at 23:12
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    $\begingroup$ @TheoBendit Oh, sorry! Didn't see that in your comment, but yea, like you said I want to just use inequalities $\endgroup$ – BeastCoder2 Apr 8 at 23:13
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    $\begingroup$ Did you mean $3-|x\color{red}{+}2|$ ? $\endgroup$ – J. W. Tanner Apr 8 at 23:51
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You have three cases to look at:

• Case 1: $x<-1$

• Case 2: $-1\le x<2$

• Case 3: $x\ge 2$

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  • $\begingroup$ @HANDMINSOUMARE Could you please give an explanation as to how you got those cases $\endgroup$ – BeastCoder2 Apr 8 at 23:07
  • $\begingroup$ Look at the roots of $x+1$ and $x-2$. $\endgroup$ – HAMIDINE SOUMARE Apr 8 at 23:10
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The first term $|x+1|$ is the distance from $x$ to $-1$, and the second term $|x-2|$ is the distance from $x$ to $2$. Now, considering that the points $-1$ and $2$ lie precisely $3$ steps apart on the number line, can you deduce where on the the number line $x$ must lie in order for the sum of those two distances to be $3$?

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Case 1: $x+1+x+2=3$ >> $x=0$

Case 2: $-(x+1)+x+2=3$ >> No Solutions

Case 3: $x+1-(x+2)=3$ >> No Solutions

Case 4: $-(x+1)-(x+2)=3$ >> $x=-3$

Therefore the two answers are: $x=0$ and $x=-3$

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  • $\begingroup$ You seem to have mixed up $x - 2$ and $x + 2$ here. $\endgroup$ – Theo Bendit Apr 8 at 23:26
  • $\begingroup$ @TheoBendit I fixed it! $\endgroup$ – BeastCoder2 Apr 8 at 23:33
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    $\begingroup$ Oh, so it was the original question that was wrong? The answerers will not be pleased. :-( $\endgroup$ – Theo Bendit Apr 8 at 23:44
  • $\begingroup$ You should also check the feasibility of your solutions by plugging them back into the original equation. Some solutions can be false in such circumstances. $\endgroup$ – Theo Bendit Apr 8 at 23:46
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You asked for solutions to $$|x+1|+|x+2|=3.$$

If $x<-2$ then $x+1$ and $x+2 <0$, so this means $-(x+1)-(x+2)=3,$ i.e., $x=-3.$

If $x\ge-1$ then $x+1$ and $x+2 \ge 0$, so this means $(x+1)+(x+2)=3,$ i.e., $x=0$.

If $-2\le x<-1$ then $x+1<0$ but $x+2\ge0$, so $-(x+1)+(x+2)=3$,

which has no solution.

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Hint: if $x+1 > 0$ , $x+1 < 0$, $x+1 = 0$ and $x-2 > 0$, $x - 2 < 0$, $x-2 = 0$, etc...There are $9$ possibilities all together.

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  • $\begingroup$ How did you get those? $\endgroup$ – BeastCoder2 Apr 8 at 23:08

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