$Double absolute value equations.$

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$Double absolute value equations.$

I'm self-studying Cambridge IGCSE additional maths and I've got to the double absolute value equations but I've been really struggling to understand how it works for a while now.

My textbook gives an example of an equation and how to solve it as shown below.

$\left | x+4 \right | +\left | x-5 \right | = 11 $

$\left | x+4 \right |= 11 -\left | x-5 \right | $

$x+4 = 11 -\left | x-5 \right |$ (1)
$x+4 = \left | x-5 \right |-11$ (2)

Using equation (1)
$\left | x-5 \right | = 7-x$
$ x-5 = 7-x$
$2x = 12$
$x=6$

Or
$x-5=-(7-x)$
$0=-2$ Which is false

Using equation (2)
$\left | x-5 \right |=x+15$
$x-5=x+15$
$0=20$ which is false

or
$ x-5=-(x+15)$
$2x=-10$
$x=-5$

The solution is $x=6$ or $x=-5$

I know how to solve an equation like

$\left | x+5 \right |= \left | x-4 \right |$

I just don't understand what is happening here. Could you help explain it step by step why each step is taken and if possible could you graph it so that I can visually see what is happening?

Thanks for the help.

Algebra

Math Gnome

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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 $x=-5$.

Mathe

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$Double absolute value equations.$

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