Question 333582
{{{1/6}}}{{{abs(4x-8)}}}{{{""=""}}}{{{6x}}}
<pre><b>
{{{abs(4x-8)}}}{{{""=""}}}{{{36x}}}

{{{4x-8}}}{{{""=""}}}{{{36x}}};  {{{4x-8}}}{{{""=""}}}{{{-36x}}}

{{{-8}}}{{{""=""}}}{{{32x}}};  {{{-8}}}{{{""=""}}}{{{-40x}}}

{{{-8/32}}}{{{""=""}}}{{{x}}};  {{{-8/(-40)}}}{{{""=""}}}{{{x}}}

{{{-1/4}}}{{{""=""}}}{{{x}}};  {{{1/5}}}{{{""=""}}}{{{x}}}  

Therefore your work is correct.

To find out if these are solutions substitute them into the 
original equation.

Substituting {{{x=-1/4}}} in the original

{{{1/6}}}{{{abs(4x-8)}}}{{{""=""}}}{{{6x}}}

{{{1/6}}}{{{abs(4(-1/4)-8)}}}{{{""=""}}}{{{6(-1/4)}}}

{{{1/6}}}{{{abs(-1-8)}}}{{{""=""}}}{{{-6/4}}}

{{{1/6}}}{{{abs(-9)}}}{{{""=""}}}{{{-3/2}}}

{{{1/6}}}{{{(9)}}}{{{""=""}}}{{{-3/2}}}

{{{9/6}}}{{{""=""}}}{{{-3/2}}}

{{{3/2}}}{{{""=""}}}{{{-3/2}}}

Certainly that's false, so {{{-1/4}}} is an extraneous solution,

which we discard.

Substituting {{{x=1/5}}} in the original

{{{1/6}}}{{{abs(4x-8)}}}{{{""=""}}}{{{6x}}}

{{{1/6}}}{{{abs(4(1/5)-8)}}}{{{""=""}}}{{{6(1/5)}}}

{{{1/6}}}{{{abs(4/5-8)}}}{{{""=""}}}{{{6/5}}}

{{{1/6}}}{{{abs(4/5-40/5)}}}{{{""=""}}}{{{6/5}}}

{{{1/6}}}{{{abs(-36/5)}}}{{{""=""}}}{{{6/5}}}

{{{1/6}}}{{{(36/5)}}}{{{""=""}}}{{{6/5}}}

{{{6/5}}}{{{""=""}}}{{{6/5}}}

That one checks, so the only solution is {{{x=1/5}}}

Edwin</pre></b>