rudejohnMadmax25 rudejohnMadmax25
  • 02-10-2016
  • Mathematics
contestada

Y^2+3y-4
________
2y^2-32

Respuesta :

Panoyin
Panoyin Panoyin
  • 02-10-2016
[tex]\frac{y^{2} + 3y - 4}{2y^{2} - 32} = \frac{y^{2} + 4y - y - 4}{2(y^{2}) - 2(16)} = \frac{y(y) + y(4) - 1(y) - 1(4)}{2(y^{2} - 16)} = \frac{y(y + 4) - 1(y - 4)}{2(y(y + 4) - 4(y + 4))} = \frac{(y + 1)(y - 4)}{2(y - 4)(y + 4)} = \frac{y + 1}{2(y + 4)} = \frac{y + 1}{2y + 8}[/tex]
Answer Link
toporc
toporc toporc
  • 02-10-2016
[tex]\frac{y^{2}+3y-4}{2y^{2}-32}=\frac{(y+4)(y-1)}{2(y^{2}-16)}=\frac{y-1}{2y-8}[/tex]
Answer Link

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