The question is: how do we solve the following inequation:
$$|x-1| - |x-3| \geq 5$$
For this question, I have tried to solve it like a normal absolute value problem but the answers I've were wrong so I'm quite stuck on this question. I thank you in advance for your support. :)
Let's review the definition of absolute value. $$\textrm{if}~x\geq0, |x|=x$$ $$\textrm{if}~x<0, |x|=-x$$ Since we know that we can solve a linear inequality without the absolute values, we need to get rid of the absolute values first. We know the "boarder numbers" are $x-1=0$ and $x-3=0$. Therefore, we need to split our answers into $3$ segments: $x\leq1, 1<x<3, \textrm{and}~x\geq3$.
$\textrm{If}~x\leq1, \textrm{then}~|x-1|=1-x \textrm{ and } |x-3|=3-x$ $$(1-x)-(3-x)\geq5$$ $$\boxed{-2\geq5} \textrm{ which is true for none of }x\leq1$$ $\textrm{If}~1<x<3, \textrm{then}~|x-1|=x-1 \textrm{ and } |x-3|=3-x$ $$(x-1)-(3-x)\geq5$$ $$2x-4\geq5$$ $$2x\geq9$$ $$\boxed{x\geq\frac{9}{2}} \textrm{ which is true for none of } 1<x<3$$
$\textrm{If}~x\geq3, \textrm{then}~|x-1|=x-1 \textrm{ and } |x-3|=x-3$ $$(x-1)-(x-3)\geq5$$ $$\boxed{2\geq5} \textrm{ which is true for none of }x\geq3$$
Therefore, the inequality is true for none of the real number $x$, or $x\in\emptyset$.