Question 149694

{{{abs(2x+1)>=3}}} 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}}} or {{{2x+1 >= 3}}} Break up the absolute value inequality using the given rule





Now lets focus on the first inequality  {{{2x+1 <= -3}}}



{{{2x+1<=-3}}} Start with the given inequality



{{{2x<=-3-1}}}Subtract 1 from both sides



{{{2x<=-4}}} Combine like terms on the right side



{{{x<=(-4)/(2)}}} Divide both sides by 2 to isolate x 




{{{x<=-2}}} Divide



Now lets focus on the second inequality  {{{2x+1 >= 3}}}



{{{2x+1>=3}}} Start with the given inequality



{{{2x>=3-1}}}Subtract 1 from both sides



{{{2x>=2}}} Combine like terms on the right side



{{{x>=(2)/(2)}}} Divide both sides by 2 to isolate x 




{{{x>=1}}} Divide




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Answer:


So our answer is


{{{x <= -2}}} or {{{x >= 1}}}




So the solution in interval notation is: <font size="8">(</font>*[Tex \LARGE \bf{-\infty,-2}]<font size="8">]</font> *[Tex \LARGE \cup]<font size="8">[</font>*[Tex \LARGE \bf{1,\infty}]<font size="8">)</font>





Here's the graph of the solution set


{{{drawing(500,80,-7, 6,-10, 10,
number_line( 500, -7, 6 ,-2,1),

blue(arrow(-2,0,-7,0)),
blue(arrow(-2,0.30,-7,0.30)),
blue(arrow(-2,0.15,-7,0.15)),
blue(arrow(-2,-0.15,-7,-0.15)),
blue(arrow(-2,-0.30,-7,-0.30)),


blue(arrow(1,0,6,0)),
blue(arrow(1,0.30,6,0.30)),
blue(arrow(1,0.15,6,0.15)),
blue(arrow(1,-0.15,6,-0.15)),
blue(arrow(1,-0.30,6,-0.30))


)}}}



Note:

There is a <b>closed</b> circle at {{{x=-2}}} which means that we're including that value from the solution set.



Also, there is a <b>closed</b> circle at {{{x=1}}} which means that we're including that value from the solution set.