Question 207433
{{{-2(3+x)<4x+4<8x}}} Start with the given compound inequality.



{{{-6-2x<4x+4<8x}}} Distribute




Break up the compound inequality to get:



{{{-6-2x<4x+4}}} AND {{{4x+4<8x}}}



So let's solve the first inequality {{{-6-2x<4x+4}}}



{{{-6-2x<4x+4}}} Start with the given inequality.



{{{-2x<4x+4+6}}} Add {{{6}}} to both sides.



{{{-2x-4x<4+6}}} Subtract {{{4x}}} from both sides.



{{{-6x<4+6}}} Combine like terms on the left side.



{{{-6x<10}}} Combine like terms on the right side.



{{{x>(10)/(-6)}}} Divide both sides by {{{-6}}} to isolate {{{x}}}. note: Remember, the inequality sign flips when we divide both sides by a negative number. 



{{{x>-5/3}}} Reduce.



Now let's solve the second inequality {{{4x+4<8x}}}



{{{4x+4<8x}}} Start with the given inequality.



{{{4x<8x-4}}} Subtract {{{4}}} from both sides.



{{{4x-8x<-4}}} Subtract {{{8x}}} from both sides.



{{{-4x<-4}}} Combine like terms on the left side.



{{{x>(-4)/(-4)}}} Divide both sides by {{{-4}}} to isolate {{{x}}}. note: Remember, the inequality sign flips when we divide both sides by a negative number. 



{{{x>1}}} Reduce.




If we combine the results {{{x>-5/3}}} <b>and</b> {{{x>1}}} we get {{{x>1}}}



Note: we're essentially performing a set intersection between the two results.


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Answer:


So the solution is {{{x>1}}}



So the answer in set-builder notation is *[Tex \LARGE \left\{x\|x>1\right\}]




Here's the graph of the solution set


{{{drawing(500,80,-9, 11,-10, 10,
number_line( 500, -9, 11),

arrow(1,0,11,0),
arrow(1,0.30,11,0.30),
arrow(1,0.15,11,0.15),
arrow(1,-0.15,11,-0.15),
arrow(1,-0.30,11,-0.30),

circle(1,0,0.3),
circle(1,0,0.3),
circle(1,0,0.3),
circle(1,0,0.3-0.02)
)}}}