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I have this expression and my goal was to find the possible x values: Because there a two $| |$ i thought i have to make four cases! But i think i did something wrong because at the end i get for $x = [2, 0.5, 1]$! Please revise my steps! Thanks

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

1: $x < 0... and... x + 1 > 0 $

2: $x < 0... and... x + 1 < 0 $

3: $x > 0... and... x + 1 > 0 $

4: $x > 0... and... x + 1 < 0 $

So then i calculated for 2 eg:

$2x + x = 3 + [-(x + 1)]$

$x = 0.5 $

But when i try $0.5$ with my expression i get $1.5 = 4.5$ so there is something wrong! Thanks for your help

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Convince yourself that there are only three cases: $x<-1$, $-1\leq x\leq0$ and $x>0$. – Michael Hoppe Dec 13 '13 at 16:28
For case 2) you should solve $2x-x=3-(x+1)=0$ leading to $x=1$. It is not a solution because the conditions mentioned in 2) are not satisfied. The conditions under 4) cannot be satisfied, so that case can be left out. Cases where $x=0$ or $x+1=0$ do not occur in your list. – drhab Dec 13 '13 at 16:35
up vote 3 down vote accepted

You have the right idea of breaking this into cases. Once you do, simply determine the absolute value of each of the expressions given your choice of $x$, which you seem to have done, then solve for $x$. Then check to see if your $x$ value works. Notice you solved for $x$ in the second case and obtained $x=1/2$. However, case $2$ said that $x<0$. But $x=1/2$! Therefore, such a solution does not exist. So continue working through the rest of the cases and see if anything other solutions you get work. You should get only one solutions when all is said and done and the solution is.....

The only solution is $x=2$.

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Thanks for your help! I only have one more little question! In case one i get $2$ but the case was: 1: $x < 0... and... x + 1 > 0 $ so $2 < 0$ ? What do i not understand? Thanks – John Smith Dec 13 '13 at 16:40
@user2724695 So $x=2$ does not work in case $1$. However, have you looked at case $3$?..... Be sure to work through all the cases! – mathematics2x2life Dec 13 '13 at 16:41
I feel so dummy but in case three i get $2x-x = 3 + x + 1$ so $0 = 4$ – John Smith Dec 13 '13 at 16:45
Case $3$: $x>0$ so $|x|=x$ and $x+1>0$ so $|x+1|=x+1$. Therefore, the equation is $2x+|x|=2x+x=3x$ and $3+|x+1|=3+x+1=x+4$. Setting these equal yields $3x=x+4$ so that $2x=4$ and $x=2$. – mathematics2x2life Dec 13 '13 at 16:47
@user2724695 That is not a case. How can $x=0$ AND $x+1=0$? That is impossible. – mathematics2x2life Dec 13 '13 at 16:52

Your fourth possibility, that $x \gt 0$ and $x+1 \lt 0$, is impossible as the two conditions are inconsistent. Other than that you should be fine. You do need to check after you get each solution that it is consistent with the set of conditions you assumed for it.

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$$2x+|x|=3+|x+1|\iff2x-3=|x+1|-|x|\iff2x-3=\pm1\iff x=\frac{3\pm1}2$$

$\iff x=1$ and $x=2$ . Now all that's left to do is to verify the two solutions.

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$x=1$ doesn't work here. After you find the solutions using this method, you should check them in the equality itself to see if all of the answers are true. Because your proof is necessary but may not be sufficient. – CODE Dec 13 '13 at 16:46
@CODE: I should check them, or the OP ? ;-) – Lucian Dec 13 '13 at 16:50
You should check them, as long as you are posting it as an answer for the OP. – CODE Dec 13 '13 at 16:51
@CODE: Is it our purpose to guide and to educate, or to do someone's homework for them ? – Lucian Dec 13 '13 at 17:00
Ofcourse, But for that purpose you should give out hints. Because if you don't, the OP may think of your answer as the main answer. – CODE Dec 13 '13 at 18:14

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