How do you simplify #(4x+16)/x div (x+4)#?

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Answer 1 · 2018-03-21T20:13:07

#4/x#

This is the same as #(4x+16)/x xx1/(x+4)# giving

#(4x+16)/(x(x+4))#

Factor out 4 from the top giving:

#(4(x+4))/(x(x+4)) color(white)("d") ->color(white)("d")4/x xx (x+4)/(x+4) #

But #(x+4)/(x+4)=1# giving

# color(white)("ddddddddd") -> color(white)("d")4/x color(white)("d") xx color(white)("d")1 = 4/x#

Answer 2 · 2018-03-21T20:15:14

#4/x#

With algebraic fractions,a good place to start is to factorise if possible.

#(4x+16)/x div (x+4)#

#=(4(x+4))/x xx 1/((x+4))" "larr# multiply and invert

Now you can cancel the factors - there is a multiplication sign between them.

#=(4cancel((x+4)))/x xx 1/cancel((x+4))#

#=4/x#