Question 158106
{{{6abs(2x + 5)=6x + 24}}}
<pre><font size = 4 color = "indigo"><b>
What is between the absolute value bars,
{{{2x+5}}}, can either be negative or 
non-negative, so we must always make two 
separate equations, one for each case:

Case 1:   
{{{2x+5>=0}}}

Then {{{2x+5}}} is non-negative and
we can just replace the absolute value bars
by parentheses, so

{{{6abs(2x + 5)=6x + 24}}}

becomes simply:

{{{6(2x + 5)=6x + 24}}}
{{{12x+30=6x+24}}}
{{{6x=-6}}}
{{{x=-1}}}

Case 2:   
{{{2x+5<0}}}

Then {{{2x+5}}} is negative and so we
must multiply what is between the absolute
value bars by -1 to make it a positive
number, so w replace {{{abs(3x+5)}}} by
{{{(-1)(2x+5)}}}.  So

{{{6abs(2x + 5)=6x + 24}}}

becomes {{{6(-1)(2x+5)=6x+24}}}

{{{-6(2x + 5)=6x + 24}}}
{{{-12x-30=6x+24}}}
{{{-18x=54}}}
{{{x=-3}}}

But we must check both solutions because
sometimes we get extraneous solutions,
that is, bogus solutions, that do not check,
and these must be discarded.

Checking {{{x=-1}}} in original:

{{{6abs(2x + 5)=6x + 24}}}
{{{6abs(2(-1) + 5)=6(-1) + 24}}}
{{{6abs(-2 + 5)=-6 + 24}}}
{{{6abs(3)=18}}}
{{{6(3)=18}}}
{{{18=18}}}

So {{{-1}}} is a solution.

Checking {{{x=-3}}} in original:

{{{6abs(2x + 5)=6x + 24}}}
{{{6abs(2(-3) + 5)=6(-3) + 24}}}
{{{6abs(-6 + 5)=-18 + 24}}}
{{{6abs(-1)=6}}}
{{{6(1)=6}}}
{{{6=6}}}

So {{{-3}}} is also a solution.

The solutions are -1 and -3.

Edwin</pre>