How do you solve #(3x-1)/x<=-1# using a sign chart?

1 Answer
Sep 12, 2016

Answer:

# x in (0,1/4]#

Explanation:

The first thing you have to do is to rewrite the disequation in a form like:

#(N(x))/(D(x))<=0#

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

Now you could find the values of #x# where:

#D(x)<0# (not #<=0# because a denominator could never be #0#)
#N(x)<=0#

#:.4x-1<=0 =>4x<=+1 =>x<=1/4#

#:.x<0#

Now you can draw the sign chart: where the single disequation are satisfied use a continuos line, else a discontinuos line. Where both the lines will be continuos or discontinuos the sign is #+# else is #-#

The sign of disequation is #<=# therefore you have to chose the #-# interval

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#:. x in (0,1/4]#