Question 158598
Given : {{{abs(2x-1)/abs(3)= 5}}}
{{{abs(2x-1)= 5 * abs(3)}}}
{{{abs(2x-1)= 15}}}

So when have 2 solutions:
{{{2x-1= 15}}}  and {{{2x-1= -15}}}
{{{2x = 16}}}   and {{{2x = -14}}}
{{{x = 8}}}   and {{{x = -7}}}

get it?