3/(y^2-9) + 4/(y+3) =1? P.S. I don't know how to do this can you please give me a answer right now when you are finished this problem?

Mar 17, 2017

$y = 0$ and $y = 4$

Explanation:

The equation

$\frac{3}{\left(y - 3\right) \left(y + 3\right)} + \frac{4}{y + 3} = 1$ can be solved first by multiplying both equation sides by $y + 3$. After that we obtain

$\frac{3}{y - 3} + 4 = \left(y + 3\right)$

multiplying both sides by $\left(y - 3\right)$ we obtain

$3 + 4 \left(y - 3\right) = {y}^{2} - 9$ or

${y}^{2} - 4 y = 0$ or

$y \left(y - 4\right) = 0$ giving the solutions

$y = 0$ and $y = 4$