How do you combine $(y-2)/(y-4)+(2y^2-15y+12)/(y^2-16)$ ?

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Answer 1 · 2016-07-21T16:27:14

$(3y-1)/(y+4)$ .

The Expression $=(y-2)/(y-4)+(2y^2-15y+12)/((y-4)(y+4))$

The $L.C.M.$ of the $Drs.$ is $y^2-16=(y-4)(y+4)$

$:. the Exp.=((y-2)(y+4)+2y^2-15y+12)/((y-4)(y+4))$

$=((y^2+2y-8+2y^2-15y+12))/((y-4)(y+4))$

$=(3y^2-13y+4)/((y-4)(y+4))$

$=(cancel((y-4))(3y-1))/(cancel((y-4))(y+4))$

$=(3y-1)/(y+4)$ .

How do you combine (y-2)/(y-4)+(2y^2-15y+12)/(y^2-16)(y-2)/(y-4)+(2y^2-15y+12)/(y^2-16)?