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.

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