How do you simplify #(7x^4y^3)/(5xy)*(2xy^7)/(21y^5)#?

1 Answer
Feb 12, 2017

Answer:

See the entire simplification process below:

Explanation:

First, use these rules of exponents to multiply the numerators and denominators:

#a = a^color(red)(1)# and #x^color(red)(a) xx x^color(blue)(b) = x^(color(red)(a) +color(blue)(b))#

#(7x^4y^3)/(5xy) * (2xy^7)/(21y^5) -> (7x^4y^3)/(5x^1y^1) * (2x^1y^7)/(21y^5) -> (14x^(4+1)y^(3+7))/(105x^1y^(1+5)) =#

#(14x^5y^10)/(105x^1y^6)#

Next, let's factor the constants:

#(14x^5y^10)/(105x^1y^6) = ((7 xx 2)x^5y^10)/((7 xx 15)x^1y^6) = ((color(red)(cancel(color(black)(7))) xx 2)x^5y^10)/((color(red)(cancel(color(black)(7))) xx 15)x^1y^6) =#

#(2x^5y^10)/(15x^1y^6)#

Now, we can use this rule for exponents to complete the simplification:

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

#(2x^5y^10)/(15x^1y^6) = (2x^(5-1)y^(10-6))/15 = (2x^4y^4)/15# or #2/15x^4y^4#