How do you solve #(x-4)/(x-3)>0#?

1 Answer
Sep 8, 2016

Answer:

#x in {(-oo,3),(4,+oo)}#

Explanation:

The "critical" values for #x# are obviously #3# and #4#.

Create a chart for the implied ranges of #x#:
#{: (, " | ",x < 3, " | ",x = 3, " | ", 3 < x < 4, " | ", x=4, " | ",x > 4), ((x-4)/(x-3), " | ",>0, " | ","undefined", " | ",< 0, " | ",=0, " | ",>0) :}#

We can see that #(x-4)/(x-3) > 0# when #x < 3# and when #x > 4#

Here's a graph of #(x-4)/(x-3)# that might help verify this result:
graph{(x-4)/(x-3) [-2.05, 10.434, -2.5, 3.74]}