How do you solve #(2x-1)/(x+5)>=0# using a sign chart?

1 Answer
Dec 19, 2016

Answer:

The answer is #x in ] -oo,-5 [ uu [1/2, +oo[#

Explanation:

Let #f(x)=(2x-1)/(x+5)#

Let's do the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##color(white)(aaaa)##-5##color(white)(aaaa)##color(white)(aaaa)##1/2##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+5##color(white)(aaaaaa)##-##color(white)(aa)##color(white)(aaa)##∥##color(white)(aaaa)##+##color(white)(aaa)####color(white)(aa)##+#

#color(white)(aaaa)##2x-1##color(white)(aaaaaa)##-##color(white)(aa)##color(white)(aa)##∥##color(white)(aaaa)##-##color(white)(aaa)####color(white)(aa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaaa)##+##color(white)(aa)##color(white)(aa)##∥##color(white)(aaaa)##-##color(white)(aaa)####color(white)(aa)##+#

Therefore,

#f(x)>=0#, when #x in ] -oo,-5 [ uu [1/2, +oo[#

graph{y-(2x-1)/(x+5)=0 [-52, 52.07, -26, 26]}