Method to solve $|x| + |2-x| \leq x+ 1$ Even if it seems really easy, I'm struggling to solve $$|x| +|2-x |\leq x+1.$$
The book says that $ x \in [1,3] $.
I first rewrote as $x+(2-x)\leq x+1$
with $x\geq 0$ and $-x-(2-x)\leq x+1$ with $x<0$. Then I solved.
For the first, I got $1\leq x$ and for the 2nd, $-3\leq x$ 
$$x\in ]-3;0[\cup[1;+\infty[ $$
It is maybe a very stupid question, but I can't see what I did wrong? 
Thanks for your help.
 A: Note that $|x| = x$ if $x \ge 0$ and $|x| = -x$ if $x < 0$.
Similarly, $|2-x| = 2-x$ if $2 -x \ge 0$ (i.e., if $x \le 2$), and $|2-x| = x-2$ if $2-x < 0$ (i.e., if $x > 2$).
One way is to consider separately the three cases:


*

*$x > 2$ (and so $x \ge 0$ also), so $|x| = x$ and $|2-x| = x-2$.
The inequality becomes $x + x - 2 \le x + 1$.

*$0 \le x \le 2$, so $|x| = x$ and $|2-x| = 2-x$.
The inequality becomes $x + 2-x \le x + 1$.

*$x < 0$ (and so $x \le 2$ also), so $|x| = -x$ and $|2-x| = 2-x$.
The inequality becomes $-x+2-x \le x + 1$.
Technically there is a fourth combination of $x < 0$ and $x \ge 2$, but clearly this case can't exist since no number can be $ < 0$ and $\ge 2$ at the same time.
Solve the inequalities in each case, and make sure the answer you get makes sense within the restriction of the corresponding case.
For example, for the second case, solving the inequality will give us $x \ge 1$.  But remember that this is the case where $0 \le x \le 2$.  Thus from the second case we only get $1 \le x \le 2$.
Solve the first and third cases, put it all together, and you'll get $x \in [1,3]$.  Let me know if you require further assistance.
A: we can write $$|x|+|x-2|\le x+1$$
the first case: $$x\geq 2$$ then we have to solve $$x+x-2\le x+1$$
second: $$0\le x<2$$ then we have $$x-x+2\le x+1$$
last case:
$$x<0$$ then we get $$-x-x+2\le x+1$$
Can you finish?
A: The mistake you have made is that,you have taken 
$|2-x| = 2-x \forall x\ge 0$, which is not true. 
For $x \gt 2$, $|2-x|$ is $ = x-2$. 
Hence, you must take intervals as three cases, as George Law points out, 
$ (i)x\le 0$,
$(ii) 0\lt x\le2$ 
$(iii) x>2$
A: There are 3 cases: 
a) $x\leq 0$ (quite useless, because it generates a bright lower bound):
$$|x| +|2-x |\leq x+1\implies -x+ 2-x\leq x+1$$
$$-3x\leq -1$$
$$3x\geq1$$
$$x\geq\frac{1}{3}$$
b) $0<x\leq 2$:
$$|x| +|2-x |\leq x+1\implies x+2-x\leq x+1$$
$$x\geq 1$$
c) $x > 2$:
$$|x| +|2-x |\leq x+1\implies x+x-2\leq x+1$$
$$2x\leq x+3\implies x\leq 3$$
Now, you know the lower bound of $x$, from case b, $x_{min}=1$, and you know the upper bound of $x$ from case c, $x_{max}=3$. Therefore, $1\leq x\leq 3\implies x\in [1;3]$
A: A change of sign is due to continuity only possible where 
$$|x| +|2-x |= x+1\iff |2- x|=x+1-|x|.$$
Squaring the equation yields 
$$2|x|(x-1)=(x-1)(x+3).$$
Since $x=1$ is a solution we may happily square
$$2|x|=x+3$$
again leading to
$$3(x-3)(x+1)=0.$$
Now consider the truth of 
$$|x| +|2-x |\leq x+1$$
in the intervals given by the zeroes $-1$, $1$, and $3$.
