How do you simplify #(64a^2b^5)/(35b^2c^3f^4)div(12a^4b^3c)/(70abcf^2)#?

1 Answer
Jan 8, 2017

Answer:

See full explanation below:

Explanation:

First step we will take is to rewrite this expression in the form below:

#((64a^2b^5)/(35b^2c^3f^4))/((12a^4b^3c)/(70abcf^2))#

Next, we will divide these fractions using the rule:

#(color(red)(a)/color(blue)(b))/(color(green)(c)/color(purple)(d)) = (color(red)(a) xx color(purple)(d))/(color(blue)(b) xx color(green)(c))#

#((64a^2b^5) xx (70abcf^2))/((35b^2c^3f^4) xx (12a^4b^3c))#

Then, we can multiply the terms in the numerator and the terms in the denominator using the rules for exponents:

#x^color(red)(a) xx x^color(blue)(b) = x^(color(red)(a)+color(blue)(b))#

#x = x^color(red)(1)#

#(4480a^(2+1)b^(5+1)cf^2)/(420a^4b^(2+3)c^(3+1)f^4)#

#(4480a^3b^6cf^2)/(420a^4b^5c^4f^4)#

We use these rules for exponents to divide this expression:

#x^color(red)(a)/x^color(blue)(b) = x^(color(red)(a)-color(blue)(b))#

#x^color(red)(a)/x^color(blue)(b) = 1/x^(color(blue)(b)-color(red)(a))#

#(4480b^(6-5))/(420a^(4-3)c^(4-1)f^(4-2))#

#(4480b^1)/(420a^1c^3f^2)#

#(4480b)/(420ac^3f^2)#

Now, we can finish by factoring the constants:

#((140 xx 64)b)/((140 xx 3)ac^3f^2)#

#((cancel(140) xx 64)b)/((cancel(140) xx 3)ac^3f^2)#

#(64b)/(3ac^3f^2)#