algebra precalculus - 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.

Answer

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

$$|x| +|2-x |\leq x+1\implies x+2-x\leq x+1$$ $$x\geq 1$$

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, $1\leq x\leq 3\implies x\in [1;3]$