Double absolute value equations.

Let's see an example wth just one absolute value. Suppose we have 

$|x-1| = 2$

Then, the usual procedure is to consider two possible clases for $|x-1|$.

Either $x-1 \geq 0$ or $x-1 < 0$. In the first case, we replace $|x-1|$ by $x-1$:

Case 1. $|x-1| = x-1$:
$x-1 = 2$, so that $x = 3$.

In the second case, we replace $|x-1|$ by $-(x-1)$:

Case 2. $|x-1| = -(x-1)$:
$-(x-1) = 2$, so that $x = -1$.

And we have found all possible solutions to the original equation $|x-1| = 2$.

This is the same logic they used, except they had two absolute vales, and for each absolute value we must consider two possible cases.

In their problem they have:

$∣x+4∣=11-∣x-5∣$.

We start by considering cases for $|x+4|$. The first case is $|x+4| = x+4$ and the second case is $|x+4| = -(x+4)$

Case 1. $|x+4| = x+4$:
$x+4=11-∣x-5∣$.

We can rewrite this as

$∣x-5∣ =7 - x$

And now, we must consider two cases for $|x-5|$: $|x-5| =x-5$ and $|x-5| =-(x-5)$

Case 1-1. $|x-5| =x-5$:
$x-5 = 7 - x$, which leads to $x = 6$

Case 1-2. $|x-5| =-(x-5)$:
$-(x-5) = 7 - x$, which leads to $0 = 2$. This is impossible, so we don't find any solution in this subcase.

Now that we have completely explored case 1, we check case 2:

Case 1. $|x+4| = -(x+4)$:
$-(x+4)=11-∣x-5∣$.

We can rewrite this as

$∣x-5∣ =x + 15$

Once again, we must consider two cases for $|x-5|$: $|x-5| =x-5$ and $|x-5| =-(x-5)$

Case 2-1. $|x-5| =x-5$:
$x-5 = x + 15$, which leads to $0 = 20$. An impossibility. We ignore this case.

Case 2-2. $|x-5| =-(x-5)$:
$-(x-5) = x + 15$, which leads to $x = -5$. 

We have exhausted all possible cases and found two possible soltions, $x=6$ and $x=-5$.