How do you simplify #(-8/21 x^2y^3) times (-7/16xy^2)#?

1 Answer
Aug 15, 2017

See a solution process below:

Explanation:

First, rewrite the expression as:

#(-8/21 xx -7/16)(x^2 xx x)(y^3 xx y^2) =>#

#(-color(red)(cancel(color(black)(8)))/(color(blue)(cancel(color(black)(21))3)) xx -color(blue)(cancel(color(black)(7)))/(color(red)(cancel(color(black)(16)))2))(x^2 xx x)(y^3 xx y^2) =>#

#(-1/3 xx -1/2)(x^2 xx x)(y^3 xx y^2) =>#

#1/6(x^2 xx x)(y^3 xx y^2)#

Next, use this rule for exponents to rewrite the #x# term:

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

#1/6(x^2 xx x)(y^3 xx y^2) => 1/6(x^2 xx x^color(red)(1))(y^3 xx y^2)#

Now, use this rule of exponents to simplify the #x# and #y# terms:

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

#1/6(x^color(red)(2) xx x^color(blue)(1))(y^color(red)(3) xx y^color(blue)(2)) =>#

#1/6x^(color(red)(2)+color(blue)(1))y^(color(red)(3)+color(blue)(2)) =>#

#1/6x^3y^5#

Or

#(x^3y^5)/6#