Question 196829
{{{abs(4x)>6}}} Start with the given inequality



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


{{{4x < -6}}} or {{{4x > 6}}} Break up the absolute value inequality using the given rule





Now lets focus on the first inequality  {{{4x < -6}}}



{{{4x<-6}}} Start with the given inequality



{{{x<(-6)/(4)}}} Divide both sides by 4 to isolate x 




{{{x<-3/2}}} Reduce



Now lets focus on the second inequality  {{{4x > 6}}}



{{{4x>6}}} Start with the given inequality



{{{x>(6)/(4)}}} Divide both sides by 4 to isolate x 




{{{x>3/2}}} Reduce




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Answer:


So our answer is


{{{x < -3/2}}} or {{{x > 3/2}}}



which looks like this in interval notation



*[Tex \LARGE \left(-\infty,-\frac{3}{2}\right)\cup\left(\frac{3}{2},\infty\right)]



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


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

blue(arrow(-2,-7,-10,-7)),
blue(arrow(-2,-6.5,-10,-6.5)),
blue(arrow(-2,-6,-10,-6)),
blue(arrow(-2,-5.5,-10,-5.5)),
blue(arrow(-2,-5,-10,-5)),
blue(arrow(2,-7,10,-7)),
blue(arrow(2,-6.5,10,-6.5)),
blue(arrow(2,-6,10,-6)),
blue(arrow(2,-5.5,10,-5.5)),
blue(arrow(2,-5,10,-5)),

circle(-1.5,-5.8,0.35),
circle(-1.5,-5.8,0.4),
circle(-1.5,-5.8,0.45),


circle(1.5,-5.8,0.35),
circle(1.5,-5.8,0.4),
circle(1.5,-5.8,0.45)




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