Question 119825

{{{abs((1/2)x-1)<1}}} Start with the given inequality



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


{{{(1/2)x-1 > -1}}} and {{{(1/2)x-1 < 1}}} Break up the absolute value inequality using the given rule



{{{-1 < (1/2)x-1 < 1}}} Combine the two inequalities to get a compound inequality




{{{0 < (1/2)x < 2}}} Add 1 to  all sides




{{{0 < x < 4}}} Multiply all sides by 2 to isolate x



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Answer:


So our answer is


{{{0 < x < 4}}}




which looks like this in interval notation



*[Tex \LARGE \left(0,4\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(line(-1.5,-7,1.65,-7)),
blue(line(-1.5,-6,1.65,-6)),
blue(line(-1.5,-5,1.65,-5)),

circle(-2,-5.8,0.35),
circle(-2,-5.8,0.4),
circle(-2,-5.8,0.45),


circle(2,-5.8,0.35),
circle(2,-5.8,0.4),
circle(2,-5.8,0.45)




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