Question 1006237
Ambiguous inequality, maybe really meant as 4/(x-1)>3/x,


{{{4/(x-1)-3/x>0}}}


{{{(4/(x-1)-3/x)(x(x-1))>0(x(x-1))}}}


{{{4x-3(x-1)>0}}}


{{{4x-3x+3>0}}}


{{{x+3>0}}}


{{{highlight(x>-3)}}} which is the interval notation form,  (-3, infinity).
Be aware, a critical value is x at 0.  The inequality will be UNDEFINED for x=0.
Another critical value is x at 1; the inequality is UNDEFINED for x=1.




You will make better sense of the choices given if your inequality really is exactly as it was shown in your question:   {{{4/x-1>3/x}}}, and the only critical value would be x at 0 being the undefined value for x in the inequality.
{{{4/x-1-3/x>0}}}
{{{1/x-1>0}}}
{{{1-1x>0}}}
{{{1-x>0}}}.
Now the critical values of x are  0 and 1.
The intervals on x to check are (-infinity,0), (0,1), and (1, infinity).


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One should get all expressions onto one side with 0 on the other side before further simplifying because the denominators may be positive OR negative, affecting the order when performing the multiplication for the order relationship.