How do you simplify the expression #(5ab^2 * 12ab)/(6ab)#?

2 Answers
Mar 20, 2018

Answer:

#10ab^2#

Explanation:

We start with:

#=>(5ab^2 * 12ab)/(6ab)#

Identify like-terms:

#=>(color(blue)(5)color(red)(a)color(orange)(b^2) * color(blue)(12)color(red)(a)color(orange)(b))/(color(blue)(6)color(red)(a)color(orange)(b))#

Let's multiply like-terms in the numerator first:

#=>((color(blue)(5)*color(blue)(12))(color(red)(a)*color(red)(a))(color(orange)(b^2)*color(orange)(b)))/(color(blue)(6)color(red)(a)color(orange)(b))#

#=>(color(blue)(60)color(red)(a^2)color(orange)(b^3))/(color(blue)(6)color(red)(a)color(orange)(b))#

Now we'll divide like-terms:

#=>color(blue)(60/6)color(red)(a^2/a)color(orange)(b^3/b)#

#=> color(green)(10ab^2)#

Mar 20, 2018

Answer:

You must follow the rules, which include multiplying exponents as you would add, and dividing as you would subtract. Your final answer should be #10ab^2#. This is how you do it:

Explanation:

#(5ab^2*12ab)/(6ab)#
You can do this 2 different ways, by multiplying across the top first or by dividing.

By multiplying first:

#(60a^2b^3)/(6ab)#
#a*a# is #a^2#, and #b^2*b# is #b^3#, because 2+1=3.
Now divide 60 by 6, #a^2# by #a#, and #b^3# by #b#.
#10ab^2#

By dividing:

#(5ab^2)/(6ab)=(5b)/6#, as the #a#'s cancel out (1-1=0).

#(5b)/6*12ab=10ab^2#.