Question 337765

{{{abs(8x-1)-1>2}}} Start with the given inequality



{{{abs(8x-1)>3}}} Add 1 to both sides.



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


{{{8x-1 < -3}}} or {{{8x-1 > 3}}} Break up the absolute value inequality using the given rule





Now lets focus on the first inequality  {{{8x-1 < -3}}}



{{{8x-1<-3}}} Start with the given inequality



{{{8x<-3+1}}}Add 1 to both sides



{{{8x<-2}}} Combine like terms on the right side



{{{x<(-2)/(8)}}} Divide both sides by 8 to isolate x 




{{{x<-1/4}}} Reduce



Now lets focus on the second inequality  {{{8x-1 > 3}}}



{{{8x-1>3}}} Start with the given inequality



{{{8x>3+1}}}Add 1 to both sides



{{{8x>4}}} Combine like terms on the right side



{{{x>(4)/(8)}}} Divide both sides by 8 to isolate x 




{{{x>1/2}}} Reduce




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Answer:


So our answer is


{{{x < -1/4}}} or {{{x > 1/2}}}



which looks like this in interval notation



*[Tex \LARGE \left(-\infty,-\frac{1}{4}\right)\cup\left(\frac{1}{2},\infty\right)]



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


{{{drawing(500,50,-10,10,-10,10,
number_line( 500, -9.875, 10.125),

blue(arrow(-0.875,-7,-10,-7)),
blue(arrow(-0.875,-6.5,-10,-6.5)),
blue(arrow(-0.875,-6,-10,-6)),
blue(arrow(-0.875,-5.5,-10,-5.5)),
blue(arrow(-0.875,-5,-10,-5)),
blue(arrow(0.875,-7,10,-7)),
blue(arrow(0.875,-6.5,10,-6.5)),
blue(arrow(0.875,-6,10,-6)),
blue(arrow(0.875,-5.5,10,-5.5)),
blue(arrow(0.875,-5,10,-5)),

circle(-0.375,-5.8,0.35),
circle(-0.375,-5.8,0.4),
circle(-0.375,-5.8,0.45),


circle(0.375,-5.8,0.35),
circle(0.375,-5.8,0.4),
circle(0.375,-5.8,0.45)




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




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Jim