How do you solve #(3/x + 2/x) >( 1- 4/x)#?

1 Answer
Jul 1, 2017

Answer:

See below.

Explanation:

Before we do anything, we must remember that #x# cannot be zero, as it causes the denominator of the fractions to be zero.

#(3/x + 2/x) =5/x#

#5/x > 1- 4/x#

#9/x > 1#

#9> x#, where #x\ne0# (or #(-oo,0) \uu (0,9)#).