# How do you solve (2y) / (2y+6) + (9y-12) / (3y+9) = (9y+11) / (y+3)?

Oct 19, 2015

Some of these fractions can be simplified.

#### Explanation:

$\frac{2 y}{2 y + 6} : \frac{2}{2} = \frac{y}{y + 3}$

$\frac{9 y - 12}{3 y + 9} : \frac{3}{3} = \frac{3 y - 4}{y + 3}$

So now we have:
$\frac{y}{y + 3} + \frac{3 y - 4}{y + 3} = \frac{9 y + 11}{y + 3}$

Since the numerators are all equal we may leave them out, on the condition $y \ne - 3$ for this would make the numerators $= 0$ and the whole equation would be senseless.

$y + 3 y - 4 = 9 y + 11 \to$

All the $y$'s to one side, all the numbers to the other:

$- 5 y = 15 \to y = - 3$

And this is exactly the solution that was forbidden.