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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. :)

user0102
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allan
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    Write your inequation in each of the cases $x<1$ , $x>3$ and $ 1\le x\le 3$. – hamam_Abdallah May 06 '21 at 00:50
  • Just to add to hamam_Abdallah's response, the reason you want to look at those ranges is because of where the absolute values have turning points. For example, you can think of $|x-1|$ as $x-1$ when $x\ge 1$, and $-(x-1)$ when $x<1$. – Amaan M May 06 '21 at 01:00
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    You're looking for places on the number line whose distance from $1$ is at least five units larger than their distance from $3$ ... – Troposphere May 06 '21 at 01:01
  • I'm not that sure you did it wrong. There is no solution; that is, the solution set is the empty set. – Steven Alexis Gregory May 06 '21 at 01:16
  • If you expect people to put in the effort to answer a question, then it is polite to put in some effort yourself. In particular, don't just say "I tried to solve it like a normal absolute value problem", but give your working. (For further feedback/help with asking questions, you can ask here.) – user1729 May 12 '21 at 09:03

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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 $x\in\emptyset$.

Interstigation
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