Question 208131


{{{abs(3x-1)<=8}}} Start with the given inequality



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


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



{{{-8 <= 3x-1 <= 8}}} Combine the two inequalities to get a compound inequality




{{{-7 <= 3x <= 9}}} Add 1 to  all sides



{{{-7/3 <= x <= 3}}}  Divide all sides by 3 to isolate x




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Answer:


So our answer is


{{{-7/3 <= x <= 3}}}



So the answer in interval notation is   <font size="8">[</font>*[Tex \LARGE \bf{-\frac{7}{3},3}]<font size="8">]</font>



Also, the answer in set-builder notation is  *[Tex \LARGE \left\{x\|-\frac{7}{3} \le x \le 3\right\}]



Here's the graph of the solution set on a number line:


{{{drawing(500,80,-12, 8,-10, 10,
number_line( 500, -12, 8 ,-7/3,3),

blue(line(-7/3,0,3,0)),
blue(line(-7/3,0.30,3,0.30)),
blue(line(-7/3,0.15,3,0.15)),
blue(line(-7/3,-0.15,3,-0.15)),
blue(line(-7/3,-0.30,3,-0.30))

)}}} Graph of the solution set


Note:

There is a <b>closed</b> circle at {{{x=-7/3}}} which means that we're including this value in the solution set

Also, there is a <b>closed</b> circle at {{{x=3}}} which means that we're including this value in the solution set.