Question 491119
<pre>
|4x+4| = 8x+16

The absolute value is already isolated so we can split into two
equations without absolute value bars:

1. One equation is when what is between the absolute value bars equals
to the right side:

 4x+4 = 8x+16
4x-8x = 16-4
  -4x = 12
    x = {{{12/(-4)}}}
    x = -3  

But we must check for extraneous solutions:

   |4x+4| = 8x+16
|4(-3)+4| = 8(-3)+16
   |12+4| = -24+1
     |16| = -23
       16 = -23

That is false so -3 is extraneous, and is tot a solution

2. Another equation is when what is between the absolute value bars equals
to -1 times the right side:

4x+4 = -1(8x+16)
4x+4 = -8x-16
 12x = -20
   x = {{{(-20)/12}}}
   x = {{{-5/3}}}

Checking for extraneous solutions:

  |4x+4| = 8x+16
|4({{{-5/3}}}+4| = 8({{{-5/3}}})+16
|{{{-20/3}}}+{{{12/3}}}| = {{{-40/3}}}+{{{48/3}}}
    |{{{-8/3}}}| = {{{8/3}}}  
       {{{8/3}}} = {{{8/3}}}
That's true so there is just one solution, {{{-5/3}}}


Edwin</pre>