How do you solve #x / 3 - 2 / 3 = 1 / x# and find any extraneous solutions?

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Answer 1 · 2017-02-19T23:07:09

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

Given:

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

Multiply through by #3x# to get:

#x^2-2x=3#

Note that this can only introduce an extraneous solution if the resulting equation has #x=0# as a root.

Subtract #3# from both sides to get:

#0 = x^2-2x-3 = (x-3)(x+1)#

Hence:

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

These are both valid solutions of the original equation.