# Inequality involving absolute values.

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.

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 $$x=8$$

$$|x-2|+|x-8|\le x-2$$ 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.

We see that $$x\geq2$$.

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