Question 151740

Let's solve the first inequality {{{-3x<-6}}}:



{{{-3x<-6}}} Start with the first inequality.



{{{x>(-6)/(-3)}}} Divide both sides by {{{-3}}} to isolate {{{x}}}. note: Remember, the inequality sign flips when we divide both sides by a negative number. 



{{{x>2}}} Reduce.



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



Now let's solve the second inequality {{{x+5<-2}}}:



{{{x+5<-2}}} Start with the second inequality.



{{{x<-2-5}}} Subtract {{{5}}} from both sides.



{{{x<-7}}} Combine like terms on the right side.



So our answer is {{{x>2}}} <font size="4"><b>or</b></font>  {{{x<-7}}}





So the solution in interval notation is: <font size="8">(</font>*[Tex \LARGE \bf{-\infty,-7}]<font size="8">)</font> *[Tex \LARGE \cup]<font size="8">(</font>*[Tex \LARGE \bf{2,\infty}]<font size="8">)</font>



So the solution in set notation is: *[Tex \LARGE \left\{x\|x<-7 \textrm{..or..} x>2\right\}]





Here's the graph of the solution set


{{{drawing(500,80,-12, 7,-10, 10,
number_line( 500, -12, 7 ),


circle(-7,0,0.25),
circle(-7,0,0.20),


blue(arrow(-7,0,-12,0)),
blue(arrow(-7,0.30,-12,0.30)),
blue(arrow(-7,0.15,-12,0.15)),
blue(arrow(-7,-0.15,-12,-0.15)),
blue(arrow(-7,-0.30,-12,-0.30)),



circle(2,0,0.25),
circle(2,0,0.20),


blue(arrow(2,0,7,0)),
blue(arrow(2,0.30,7,0.30)),
blue(arrow(2,0.15,7,0.15)),
blue(arrow(2,-0.15,7,-0.15)),
blue(arrow(2,-0.30,7,-0.30))


)}}}



Note:

There is an <b>open</b> circle at {{{x=-7}}} which means that we're excluding that value from the solution set.



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