Respuesta :

[tex] \frac{4}{y^{2}-9 } + \frac{5}{y+3} = \frac{4}{(y-3)(y+3) } + \frac{5}{y+3}= \frac{4}{(y-3)(y+3) } + \frac{5(y-3)}{(y-3)(y+3)}= \\ \\=\frac{4}{(y-3)(y+3) } + \frac{5y-15}{(y-3)(y+3)}= \frac{5y-11}{(y-3)(y+3)}=\frac{5y-11}{y^{2}-9} [/tex]

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