How do you simplify #(2y-6)/(3y^2-y^3)#?

2 Answers
Alan P.
Dec 22, 2016

#color(green)(-2/y^2)#

Explanation:

Note that #3y^2-y^3=-y^3+3y^2#

#color(white)("XXX")=(-y^2)(y-3)#

#color(white)("XXX")=(-y^2)(((2y-6))/2)#

#color(white)("XXX")=(-(y^2)/2)(2y-6)#

Therefore
#color(white)("XXX")(2y-6)/(-y^3+3y^2)=(cancel(2y-6))/(-y^2/2(cancel(2y-6)))=-2/y^2#

EZ as pi
Dec 22, 2016

#=-2/y^2#

Explanation:

#(2y-6)/(3y^2-y^3)" "# factorise as far as possible

#=(2(y-3))/(y^2(3-y))#

Notice that the brackets are the same except for the signs.
Do a switch round:

#color(red)((3-y) = -(-3+y) = -(y-3))#

#=(2(y-3))/(y^2color(red)((3-y))#

#=(2(y-3))/(-y^2color(red)((y-3))#

#=(2cancel((y-3)))/(-y^2cancel((y-3))#

#=-2/y^2#

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