Double absolute value problem

Question

How do I even go about solving something like this:

$|x-1| - |x-3| \geq 5$

I know that if the question is this:

$|x-1| > 5$

I can split that up into:

$|x-1| > 5$

and

$|x-1| < -5$

But I'm a little confused about this problem.

Answer 1

You are right to be suspicious:

$|x-1| - |x-3| \ge 5 \implies$

$|x - 1| \ge 5 + |x-3| \ge 5$ so

$x -1 \ge 5$ or $x - 1 \le -5$ so $x \ge 6$ or $x \le -4$.

But if $x \ge 6$ then $x > 3$ and $|x-1| = x-1$ and $|x-3| = x-3$ so

$(x-1) - (x-3) = 2 \ge 5$ which is clearly false.

And if $x \le -4$ then $x-1 < 0$ and $x - 3< 0$ so

$|x-1| - |x-3| = -(x-1)- (-(x-3)) = -x + 1 + x -3 = -3 \ge 5$ which is even more obviously false.

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Well, you can always do four choises:

1)$(x-1) \ge 0$ and $x-3 \ge 0$ so $x \ge 1$ and $x \ge 3$

So $x\ge 3$ and $|x-1| - |x-3| = (x-1) -(x-3) = -1+3 = 2 \ge 5$ That's impossible.

2)$x-1 < 0$ and $x \ge 0$ so $x < 1$ and $x \ge 3$.

That is impossible.

3) $x-1 \ge 0$ and $x - 3 < 0$ so $x \ge 1$ and $x < 3$ so $1 \le x <3$.

So $|x-1| - |x-3| = (x-1) - (3-x) = 2x -4 \ge 5$ so $2x \ge 9$ $x \ge 4.5$.

So $1 \le x < 3$ and $x \ge 4.5$.

That is ismpossible.

Or

4) $x-1 < 0$ and $x -3 < 0$ so $x < 1$ and $x < 3$ so $x < 1$

So $|x-1| - |x-3| = (1-x) - (3-x) = -2 \ge 5$.

That is impossible.

There is not solution.

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Then there is the triangle inequality..

$|a+b| \le |a| + |b|$ so

$|a + b| - |b| \le |a|$. Let $a+b= c$

$|c| - |b| \le |c-b|$

So $|x-1| - |x-3| \le |(x-1) -(x-3)| = |2| = 2$.

So $|x-1| - |x-3| \ge 5 > 2$ is impossible.

Answer 2

You need to partition the real line into segments, such that for each segemnt the signs of $x-1$ and $x-3$ stay unchanged. For instance, when $x\in (\infty,1)$ the inequality becomes

$$-(x-1)-(-(x-3))\geq5$$

it can be simplified to

$$-2\geq5$$

which is never true.

So, you do the same for all the partitions $(a,b)$ and if you find some solutions for $x$, you take the intersection of the solution with $(a,b)$, because the simplification of the main inequality is only valid on $(a,b)$.