Question 151735

Let's solve the first inequality {{{x+5>2x+1}}}:



{{{x+5>2x+1}}} Start with the first inequality.



{{{x>2x+1-5}}} Subtract {{{5}}} from both sides.



{{{x-2x>1-5}}} Subtract {{{2x}}} from both sides.



{{{-x>1-5}}} Combine like terms on the left side.



{{{-x>-4}}} Combine like terms on the right side.



{{{x<(-4)/(-1)}}} Divide both sides by {{{-1}}} to isolate {{{x}}}. note: Remember, the inequality sign flips when we divide both sides by a negative number. 



{{{x<4}}} Reduce.



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



{{{-4x<-8}}} Start with the second inequality.



{{{x>(-8)/(-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>2}}} Reduce.



So our answer is {{{x<4}}} <font size="4"><b>and</b></font>  {{{x>2}}}



This answer combines to the compound inequality {{{2<x<4}}}




So the answer in interval notation is   <font size="8">(</font>*[Tex \LARGE \bf{2,4}]<font size="8">)</font>



So the answer in set-builder notation is  *[Tex \LARGE \left\{x\|2 < x < 4\right\}]



Here's the graph of the solution set


{{{drawing(500,80,-3, 9,-10, 10,
number_line( 500, -3, 9 ),

blue(line(2,0,4,0)),
blue(line(2,0.30,4,0.30)),
blue(line(2,0.15,4,0.15)),
blue(line(2,-0.15,4,-0.15)),
blue(line(2,-0.30,4,-0.30)),
circle(2,0,0.25),circle(2,0,0.20),
circle(4,0,0.25),
circle(4,0,0.20)

)}}} Graph of the solution set


Note:

There is an <b>open</b> circle at {{{x=2}}} which means that we're excluding this value from the solution set

Also, there is an <b>open</b> circle at {{{x=4}}} which means that we're excluding this value from the solution set.