# How do you solve 1+(3y)/(y-2)= 6/(y-2) and find any extraneous solutions?

Nov 9, 2017

There is only one solution, $y = 2$

However this is an extraneous solution because it makes the denominators equal to $0$

Hence there is no solution.

#### Explanation:

Both the denominators are the same, so re-arrange the fraction terms on the same side:

$1 = \frac{6}{y - 2} - \frac{3 y}{y - 2} \text{ } \leftarrow y \ne 2$

$1 = \frac{6 - 3 y}{y - 2} \text{ } \leftarrow$ cross-multiply

$y - 2 = 6 - 3 y$

$y + 3 y = 6 + 2$

$4 y = 8$

$y = 2$

However, this solution is not valid because it makes the denominators equal to $0$