Question 141091
{{{abs(3x-6)>3}}} Start with the given inequality



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


{{{3x-6 < -3}}} or {{{3x-6 > 3}}} Break up the absolute value inequality using the given rule





Now lets focus on the first inequality  {{{3x-6 < -3}}}



{{{3x-6<-3}}} Start with the given inequality



{{{3x<-3+6}}}Add 6 to both sides



{{{3x<3}}} Combine like terms on the right side



{{{x<(3)/(3)}}} Divide both sides by 3 to isolate x 




{{{x<1}}} Divide



Now lets focus on the second inequality  {{{3x-6 > 3}}}



{{{3x-6>3}}} Start with the given inequality



{{{3x>3+6}}}Add 6 to both sides



{{{3x>9}}} Combine like terms on the right side



{{{x>(9)/(3)}}} Divide both sides by 3 to isolate x 




{{{x>3}}} Divide




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Answer:


So our answer is


{{{x < 1}}} or {{{x > 3}}}


which in set notation is 



*[Tex \LARGE \left\{x\|x<1 \textrm{..or..} x>3\right\}]



which looks like this in interval notation



*[Tex \LARGE \left(-\infty,1\right)\cup\left(3,\infty\right)]



if you wanted to graph the solution set, you would get


{{{drawing(500,50,-10,10,-10,10,
number_line( 500, -8, 12),

blue(arrow(-1.5,-7,-10,-7)),
blue(arrow(-1.5,-6.5,-10,-6.5)),
blue(arrow(-1.5,-6,-10,-6)),
blue(arrow(-1.5,-5.5,-10,-5.5)),
blue(arrow(-1.5,-5,-10,-5)),
blue(arrow(1.5,-7,10,-7)),
blue(arrow(1.5,-6.5,10,-6.5)),
blue(arrow(1.5,-6,10,-6)),
blue(arrow(1.5,-5.5,10,-5.5)),
blue(arrow(1.5,-5,10,-5)),

circle(-1,-5.8,0.35),
circle(-1,-5.8,0.4),
circle(-1,-5.8,0.45),


circle(1,-5.8,0.35),
circle(1,-5.8,0.4),
circle(1,-5.8,0.45)




)}}} Graph of the solution set in blue and the excluded values represented by open circles