Question 143741
{{{abs((2x+1)/3)>5}}} Start with the given inequality



Break up the absolute value (remember, if you have {{{abs(x)>a}}}, then {{{x < -a}}} or {{{x > a}}})


{{{(2x+1)/3 < -5}}} or {{{(2x+1)/3 >5}}} Break up the absolute value inequality using the given rule




{{{cross(3)((2x+1)/cross(3)) < 3(-5)}}} or {{{cross(3)((2x+1)/cross(3)) > 3(5)}}} Multiply both sides by the LCD 3 to clear the fractions



{{{2x+1 < -15}}} or {{{2x+1 > 15}}} Distribute and multiply



{{{x< -8}}} or {{{x > 7}}} Isolate x in each case




So our solution is



{{{x< -8}}} or {{{x > 7}}} 




which looks like this in interval notation


<font size="8">(</font>*[Tex \LARGE \bf{-\infty,-8}]<font size="8">)</font>*[Tex \LARGE \cup]<font size="8">(</font>*[Tex \LARGE 7,\infty]<font size="8">)</font>