Question 126360
{{{6-c<2c+3<=8+c}}} Start with the given inequality




Break up the inequality. Remember {{{a<b<c}}} is equivalent to the group {{{a<b}}} and {{{b<c}}}



{{{6-c < 2c+3 }}}  and  {{{2c+3 < =8+c }}}




Now let's solve the first inequality




{{{6-c<2c+3}}} Start with the given inequality




{{{-c<2c+3-6}}}Subtract 6 from both sides



{{{-c-2c<3-6}}} Subtract 2c from both sides



{{{-3c<3-6}}} Combine like terms on the left side



{{{-3c<-3}}} Combine like terms on the right side



{{{c>(-3)/(-3)}}} Divide both sides by -3 to isolate c  (note: Remember, dividing both sides by a negative number flips the inequality sign) 




{{{c>1}}} Divide


--------------------------------------------------------------

Answer:

So the first part of our answer is {{{c>1}}} 





<hr>




{{{2c+3<=8+c}}} Start with the given inequality




{{{2c<=8+c-3}}}Subtract 3 from both sides



{{{2c-c<=8-3}}} Subtract c from both sides



{{{c<=8-3}}} Combine like terms on the left side



{{{c<=5}}} Combine like terms on the right side


--------------------------------------------------------------

Answer:

So the second part of our answer is {{{c<=5}}} 






--------------------------------------------------


Putting the the two answers together, we get:

{{{c>1}}} and {{{c<=5}}} 



Also, the two solutions form the compound inequality


{{{1<c<=5}}}






So the solution in interval notation is: (1,5]



Now let's graph the solution set


{{{drawing(500,50,-10,10,-10,10,
number_line( 500, -7, 13),
blue(line(-1.75,-7,1.85,-7)),
blue(line(-1.75,-6,1.85,-6)),
blue(line(-1.75,-5,1.85,-5)),

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

circle(2,-5.8,0.10),
circle(2,-5.8,0.15),
circle(2,-5.8,0.20),
circle(2,-5.8,0.25),
circle(2,-5.8,0.30),
circle(2,-5.8,0.35),
circle(2,-5.8,0.40)


)}}}


Note: at {{{c=1}}} there is a <font size="4"><b>open</b></font>  circle (which means this point is excluded) and at {{{c=5}}} there is a <font size="4"><b>closed</b></font> circle (which means this point is included)