Question #e97c3

1 Answer
Mar 14, 2017

Answer:

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

Explanation:

Let's simplify the equation

#(2x+1)/(x-1)<=3#

#(2x+1)/(x-1)-3<=0#

#(2x+1-3x+3)/(x-1) <=0#

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

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

We can build the sign chart

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

#color(white)(aaaa)##x-1##color(white)(aaaa)##-##color(white)(aa)##||##color(white)(aaaa)##+##color(white)(aaaa)##+#

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

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

Therefore,

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