Answer:
Step-by-step explanation:
Hello,
We know that
[tex]|x|=\begin{cases}x & \text{if } x\geq 0 \\ -x & \text{if } x<0 \end{cases}[/tex]
So we need to take into account two cases
Case 1 - [tex]x-3\geq 0<=> x\geq 3[/tex]
Then, |x-3|=x-3
||x-3|-2|=|x-3-2|=|x-5|
Either x-5 is positive and then |x-5|=x-5 and
[tex]|x-5|\leq 1<=>x-5\leq 1<=>x\leq 6[/tex]
Or x-5 is negative and then, |x-5|=-x+5
[tex]|x-5|\leq 1<=>-x+5\leq 1<=>4\leq x[/tex]
So the solution is [4;6]
Case 2 - [tex]x-3< 0<=> x< 3[/tex]
Then, |x-3|=-x+3
||x-3|-2|=|-x+3-2|=|-x+1|
Either -x+1 is positive and then |-x+1|=-x+1 and
[tex]|-x+1|\leq 1<=>-x+1\leq 1<=>0\leq x[/tex]
Or -x+1 is negative and then, |-x+1|=x-1
[tex]|-x+1|\leq 1<=>x-1\leq 1<=>x\leq 2[/tex]
So the solution is [0;2]
Conclusion
The solution is [0;2]∪[4;6]
Thanks
