Question 174309

{{{abs(4x-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}}})


{{{4x-3 < -5}}} or {{{4x-3 > 5}}} Break up the absolute value inequality using the given rule





Now lets focus on the first inequality  {{{4x-3 < -5}}}



{{{4x-3<-5}}} Start with the given inequality



{{{4x<-5+3}}}Add 3 to both sides



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



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




{{{x<-1/2}}} Reduce



Now lets focus on the second inequality  {{{4x-3 > 5}}}



{{{4x-3>5}}} Start with the given inequality



{{{4x>5+3}}}Add 3 to both sides



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



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




{{{x>2}}} Divide




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Answer:


So our answer is


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



which looks like this in interval notation



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



if you wanted to graph the solution set on a number line, you would get


{{{drawing(500,50,-10,10,-10,10,
number_line( 500, -9.25, 10.75),

blue(arrow(-1.75,-7,-10,-7)),
blue(arrow(-1.75,-6.5,-10,-6.5)),
blue(arrow(-1.75,-6,-10,-6)),
blue(arrow(-1.75,-5.5,-10,-5.5)),
blue(arrow(-1.75,-5,-10,-5)),
blue(arrow(1.75,-7,10,-7)),
blue(arrow(1.75,-6.5,10,-6.5)),
blue(arrow(1.75,-6,10,-6)),
blue(arrow(1.75,-5.5,10,-5.5)),
blue(arrow(1.75,-5,10,-5)),

circle(-1.25,-5.8,0.35),
circle(-1.25,-5.8,0.4),
circle(-1.25,-5.8,0.45),


circle(1.25,-5.8,0.35),
circle(1.25,-5.8,0.4),
circle(1.25,-5.8,0.45)




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