How do solve #(3x+1)/(x+4)<=1# and write the answer as a inequality and interval notation?

1 Answer
Mar 10, 2018

Answer:

The solution is #x in (-4,3/2]# or #-4 < x <= 3/2#

Explanation:

Let's rewrite and simplify the inequality

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

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

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

#((2x-3))/(x+4)<=0#

Let #f(x)=((2x-3))/(x+4)#

Now, we can build the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaa)##-4##color(white)(aaaaaaa)##3/2##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+4##color(white)(aaaaaa)##-##color(white)(aa)##0##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##2x-3##color(white)(aaaaa)##-##color(white)(aa)####color(white)(aaaaa)##-##color(white)(aa)##0##color(white)(aa)##+#

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

Therefore,

#f(x)<=0# when #x in (-4,3/2]# or #-4 < x <= 3/2#

graph{(3x+1)/(x+4)-1 [-27.09, 18.51, -12.22, 10.59]}