How do you solve #x/3 - 2/3 = 1/x#?

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Answer 1 · 2016-05-27T14:28:31

#x = -1# or #x = 3#

First we want to eliminate all of the denominators. To do so, multiply everything by #3x# (product of all denominators).

#3x(x/3) - 3x(2/3) = 3x(1/x)#

#x^2 - 2x = 3#

Move the #3# to the other side so that everything equals zero

#x^2 - 2x - 3 = 0#

This is factorable. We need numbers that add to #-2# and multiply to #-3#

#(x+1)(x-3) = 0#

Ergo, #x# is either #-1# or #3#. Test by subbing into the original equation

#-1/3 - 2/3 = 1/-1#

#-3/3 = 1/-1#

yes!

#3/3 - 2/3 = 1/3#

yes!