Question 191375


Let's solve the first inequality {{{4x+3<-9}}}:



{{{4x+3<-9}}} Start with the first inequality.



{{{4x<-9-3}}} Subtract {{{3}}} from both sides.



{{{4x<-12}}} Combine like terms on the right side.



{{{x<(-12)/(4)}}} Divide both sides by {{{4}}} to isolate {{{x}}}. 



{{{x<-3}}} Reduce.



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



Now let's solve the second inequality {{{6x-7>-1}}}:



{{{6x-7>-1}}} Start with the second inequality.



{{{6x>-1+7}}} Add {{{7}}} to both sides.



{{{6x>6}}} Combine like terms on the right side.



{{{x>(6)/(6)}}} Divide both sides by {{{6}}} to isolate {{{x}}}. 



{{{x>1}}} Reduce.



So our answer is {{{x<-3}}} <font size="4"><b>or</b></font>  {{{x>1}}}



So the solution in interval notation is: <font size="8">(</font>*[Tex \LARGE \bf{-\infty,-3}]<font size="8">)</font> *[Tex \LARGE \cup]<font size="8">(</font>*[Tex \LARGE \bf{1,\infty}]<font size="8">)</font>



Also, the solution in set builder notation is: *[Tex \LARGE \left\{x\|x<-3 \ \textrm{or} \ x>1\right\}]





Here's the graph of the solution set


{{{drawing(500,80,-8, 6,-10, 10,
number_line( 500, -8, 6 ),


circle(-3,0,0.25),
circle(-3,0,0.20),


blue(arrow(-3,0,-8,0)),
blue(arrow(-3,0.30,-8,0.30)),
blue(arrow(-3,0.15,-8,0.15)),
blue(arrow(-3,-0.15,-8,-0.15)),
blue(arrow(-3,-0.30,-8,-0.30)),



circle(1,0,0.25),
circle(1,0,0.20),


blue(arrow(1,0,6,0)),
blue(arrow(1,0.30,6,0.30)),
blue(arrow(1,0.15,6,0.15)),
blue(arrow(1,-0.15,6,-0.15)),
blue(arrow(1,-0.30,6,-0.30))


)}}}



Note:

There is an <b>open</b> circle at {{{x=-3}}} which means that we're excluding that value from the solution set.



Also, there is an <b>open</b> circle at {{{x=1}}} which means that we're excluding that value from the solution set.