Inequality involving absolute values.

Question

I want to ask is whether there is a method to solve following inequality more easily and compactly or it is the only method.
$$|x-2|+|x-8|\le x-2$$

What I know is taking $x<2,8>x>2,x>8$ while solving this, I have to take care whether the answer satisfies these inequalities which makes it very long and cumbersome.

Is there any other method to solve this or can I skip any of these steps or any modification to the solution making it easy. I've to solve hundreds of questions like this. Help me. I can't find anything suitable on internet.

Answer 1

Hint: You have to distinguish three cases: $$x\geq 8,2\le x<8,x<2$$ In the first case we have :$$x-2+x-8\le x-2$$ In the second one: $$x-2-x+8\le x-2$$ And in the last case: $$-x+2+x-8\le x-2$$ Can you proceed? The solution is $$x=8$$

Answer 2

$|x-2|+|x-8|\le x-2$ We have 2 cases,

1.If $|x-2|=x-2$ then you cancel this term to get $|x-8| \le 0 \implies x=8$

2.If $|x-2|=-(x-2)$ then the inequality becomes $|x-8| \le 2(x-2)$ but since in this case $x-2<0$ then also $x-8<0$ which means that $|x-8|=-(x-8)$ and you proceed..

It is nearly the same thing and I don't think there is a much easier method to do it because you have to pass through all the cases.

Answer 3

We see that $x\geq2$.

Thus, we need to solve $$|x-8|\leq0$$ which gives $x=8.$