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

1 Answer
Lucio Falabella
Sep 12, 2016

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]

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