<PRE><B>Question 1084597</B>

{{{abs(2x+1)+1&lt;=8}}}

{{{abs(2x+1)&lt;=8-1}}}...recall that {{{abs(2x+1)=sqrt(2x+1)}}}; so, you have {{{2x+1}}} and {{{-(2x+1)}}}

then,

{{{2x+1&lt;=7}}}

{{{2x&lt;=7-1}}}

{{{2x&lt;=6}}}

{{{x&lt;=6/2}}}

{{{x&lt;=3}}}

or

{{{-(2x+1)&lt;=7}}}

{{{-2x-1&lt;=7}}}

{{{-7-1&lt;=2x}}}

{{{-8&lt;=2x}}}

{{{-8/2&lt;=x}}}

{{{-4&lt;=x}}}



solution:

{{{-4&lt;=x&lt;3}}}


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