Question 528752
It is not possible because you only multiplied {{{8/8}}} which is 1, I sugguest you use this one:

{{{-5+x/8=2x}}}
{{{-5(8)+8(x/8)=2x(8)}}} --- You can multiply ALL the denominator's LCM to ALL terms, by doing this you are able to remove all the denominators.
{{{-40+x=16x}}}
{{{-15x=40}}}
{{{x=40/-15}}}
{{{x=-8/3}}}