How do you solve #12/(x+4)<=4#?

2 Answers
Jun 24, 2018

Answer:

The solution is #x in (-oo,-4)uu [-1,+oo)#

Explanation:

You cannot do crossing over.

The inequality is

#(12)/(x+4)<=4#

#<=>#, #(12)/(x+4)-4<=0#

#<=>#, #(12-4(x+4))/(x+4)<=0#

#<=>#, #(12-16-4x)/(x+4)<=0#

#<=>#, #(-4-4x)/(x+4)<=0#

#<=>#, #(4(1+x))/(x+4)>=0#

Let #f(x)=(4(1+x))/(x+4)#

Build a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-4##color(white)(aaaa)##-1##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+4##color(white)(aaaaa)##-##color(white)(aa)##||##color(white)(aa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x+1##color(white)(aaaaa)##-##color(white)(aa)####color(white)(aaa)##-##color(white)(aa)##0##color(white)(aa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##+##color(white)(aa)##||##color(white)(aa)##-##color(white)(aa)##0##color(white)(aa)##+#

Therefore,

#f(x)>=0# when #x in (-oo,-4)uu [-1,+oo)#

Jun 24, 2018

Answer:

#12/(x+4)<=4# for #x<-4# and #x>=-1#

Explanation:

As the expression is undefined for #x=-4#, we want to stay away from that value.

Before we work on the expression algebraically, let's draw a graph:
graph{-(x+1)/(x+4) [-13.21, 6.79, -5.72, 4.28]}

Based on the graph we can see that the unequality is fulfilled for
#x<-4# and #x>=-1#

Let us clean up the expression to make it easier to work with:
An equivalent expression is

#3/(x+4)<=1#

#3/(x+4)-1<=0#

#(3-(x+4))/(x+4)<=0#

#-(x+1)/(x+4)<=0#

As this is undefined for #x=-4#, we need to consider two situations: #x> -4# and #x<-4#

1) #x> -4#: As #x+4>0# we can multiply both sides with the denominator #x+4# and still keep the sign of inequality:

#-((x+1)(x+4))/(x+4)<=0#

#-(x+1)<=0#

#x+1>=0#

#x>=-1#
Therefore #12/(x+4)<=4# when #x>=-1#

2) #x< -4#: Now the denominator #(x+4)# is negative, so if we multiply the unequality with the value of denominator, we have to turn the unequal sign around:

#-((x+1)(x+4))/(x+4)>=0#

#-(x+1)>=0#

#x+1<=0#

#x<=-1#

As the starting point was that #x<-4#, this means that #12/(x+4)<=4# for all #x<-4#