Question 173123
I'm assuming that the inequality is {{{-(3*7)/(14/2)<=(-4^2-8(x-2)+7x)/(((2/3)x + 17y)^0+5)<=-(12*3)/(-3^2)}}}




First, note that ANY number (except 0) to the zeroth power is 1. So this means that {{{((2/3)x + 17y)^0=1}}}



Also, -3*7=-21 and {{{14/2=7}}}. So the left side goes from {{{
-(3*7)/(14/2)}}} to {{{-21/7}}} which reduces to {{{-3}}}



Also, -12*3=-36 and {{{-3^2=-9}}}. So the right side goes from {{{-(12*3)/(-3^2)}}} to {{{-36/(-9)}}} which reduces to {{{4}}}



So the inequality



{{{-(3*7)/(14/2)<=(-4^2-8(x-2)+7x)/(((2/3)x + 17y)^0+5)<=-(12*3)/(-3^2)}}}



simplifies to 




{{{-3<=(-4^2-8(x-2)+7x)/(1+5)<=4}}} 




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{{{-3<=(-4^2-8(x-2)+7x)/(1+5)<=4}}} Start with the given compound inequality



{{{-3<=(-16-8(x-2)+7x)/(1+5)<=4}}} Square 4 to get 16



{{{-3<=(-16-8x+16+7x)/(1+5)<=4}}} Distribute



{{{-3<=(-x)/(6)<=4}}} Combine like terms.



{{{-18<=-x<=24}}} Multiply ALL sides by 6



{{{18>=-x>=-24}}} Divide ALL sides by -1 to isolate "x" (this will flip the signs)




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Answer:



So the solution is {{{-24<=x<=18}}}





So the answer in interval notation is   <font size="8">[</font>*[Tex \LARGE \bf{-24,18}]<font size="8">]</font>



Also, the answer in set-builder notation is  *[Tex \LARGE \left\{x\|-24 \le x \le 18\right\}]



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


{{{drawing(500,80,-29, 23,-10, 10,
number_line( 500, -29, 23 ,-24,18),

blue(line(-24,0,18,0)),
blue(line(-24,0.30,18,0.30)),
blue(line(-24,0.15,18,0.15)),
blue(line(-24,-0.15,18,-0.15)),
blue(line(-24,-0.30,18,-0.30))

)}}} Graph of the solution set


Note:

There is a <b>closed</b> circle at {{{x=-24}}} which means that we're including this value in the solution set

Also, there is a <b>closed</b> circle at {{{x=18}}} which means that we're including this value in the solution set.