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

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Answer 1 · 2016-06-19T08:37:34

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

Given:

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

Multiply both sides by #3x# to get:

#(x-2)x = 3#

Expand the left hand side and subtract #3# from both sides to get:

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

Factoring the left hand side we find:

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

So:

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

You can check these values by substituting back into the original equation:

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

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