How do you solve #(2x)/(x-5)>=3#?

1 Answer
Aug 4, 2018

Answer:

The solution is #x in (5,15]#

Explanation:

You cannot do crossing over in inequalities

#(2x)/(x-5)>=3#

#<=>#, #(2x)/(x-5)-3>=0#

#<=>#, #(2x-3(x-5))/(x-5)>=0#

#<=>#, #(2x-3x+15)/(x-5)>=0#

#<=>#, #(-x+15)/(x-5)>=0#

Let #f(x)=(-x+15)/(x-5)#

Build a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaaaa)##5##color(white)(aaaaaaa)##15##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x-5##color(white)(aaaaa)##-##color(white)(aaaa)##||##color(white)(aa)##+##color(white)(aaaaaa)##+#

#color(white)(aaaa)##-x+15##color(white)(aaa)##+##color(white)(aaaaa)####color(white)(aa)##+##color(white)(aaa)##0##color(white)(aa)##-#

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

Therefore,

#f(x)>=0# when #x in (5,15]#

graph{(2x)/(x-5)-3 [-25.56, 32.17, -14.65, 14.22]}