# How do you solve 1/(6k^2)=1/(3k^2)-1/k and check for extraneous solutions?

Jul 31, 2017

$k = \frac{1}{6}$

#### Explanation:

First, let's multiply everything by the least common multiple of the denominators (which is $6 {k}^{2}$):

$\frac{6 {k}^{2}}{6 {k}^{2}} = \frac{6 {k}^{2}}{3 {k}^{2}} - \frac{6 {k}^{2}}{k}$

$1 = 2 - 6 k$

$6 k = 1$

$k = \frac{1}{6}$

The only domain restriction is $k \ne 0$, so there are no extraneous solutions:

color(blue)(ul(k = 1/6