Question 158239


{{{5*abs(2b+1)-3<=7}}} Start with the given inequality



{{{5*abs(2b+1)<=10}}} Add 3 to both sides.



{{{abs(2b+1)<=2}}} Divide both sides by 5




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


{{{2b+1 >= -2}}} and {{{2b+1 <= 2}}} Break up the absolute value inequality using the given rule



{{{-2 <= 2b+1 <= 2}}} Combine the two inequalities to get a compound inequality




{{{-3 <= 2b <= 1}}} Subtract 1 from  all sides



{{{-3/2 <= b <= 1/2}}}  Divide all sides by 2 to isolate b




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Answer:


So our answer is


{{{-3/2 <= b <= 1/2}}}




So the answer in interval notation is   <font size="8">[</font>*[Tex \LARGE \bf{-\frac{3}{2},\frac{1}{2}}]<font size="8">]</font>



and the answer in set-builder notation is  *[Tex \LARGE \left\{b\|-\frac{3}{2} \le b \le \frac{1}{2}\right\}]



Here's the graph of the solution set


{{{drawing(500,80,-8, 6,-10, 10,
number_line( 500, -8, 6 ,-3/2 , 1/2),

blue(line(-3/2 ,0, 1/2,0)),
blue(line(-3/2 ,0.30, 1/2,0.30)),
blue(line(-3/2 ,0.15, 1/2,0.15)),
blue(line(-3/2 ,-0.15, 1/2,-0.15)),
blue(line(-3/2 ,-0.30, 1/2,-0.30))

)}}} Graph of the solution set


Note:

There is a <b>closed</b> circle at {{{b=-3/2 }}} which means that we're including this value in the solution set

Also, there is a <b>closed</b> circle at {{{b= 1/2}}} which means that we're including this value in the solution set.