Question

Calcule
(y+3)/(y-3)-2/(y+3)-(24+4y)/(y²-9)

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Answer 1

$\dfrac{(y+3)}{(y-3)}-\dfrac{2}{(y+3)}-\dfrac{24+2y}{(y^2-9)}$

$y^2-9=(y+3)(y-3)$

$\dfrac{(y+3)}{(y-3)}-\dfrac{2}{(y+3)}-\dfrac{24+2y}{(y^2-9)}=\dfrac{(y+3)}{(y-3)}-\dfrac{2}{(y+3)}-\dfrac{24+2y}{(y+3)(y-3)}$

$\dfrac{(y+3)}{(y-3)}-\dfrac{2}{(y+3)}-\dfrac{24+2y}{(y+3)(y-3)}=\dfrac{(y+3)(y+3)-2(y-3)-24-4y}{(y+3)(y-3)}$

$\dfrac{(y+3)}{(y-3)}-\dfrac{2}{(y+3)}-\dfrac{24+2y}{(y+3)(y-3)}=\dfrac{y^2+6y+9-2y+6-24-4y}{(y+3)(y-3)}$

$\dfrac{(y+3)}{(y-3)}-\dfrac{2}{(y+3)}-\dfrac{24+2y}{(y+3)(y-3)}=\dfrac{y^2-8}{(y-3)(y+3)}$