How do you divide #(24x^2y^2) / (14x^3y^7)#?

2 Answers
Jul 9, 2015

Answer:

#(24x^2y^2)/(14x^3y^7) = 12/(7xy^5)#

Explanation:

#(24x^2y^2)/(14x^3y^7)#

#color(white)("XXXX")##=color(red)((24/14)) * color(blue)(((x^2)/(x^3))) * color(green)(((y^2)/(y^7)))#

#color(white)("XXXX")##= color(red)((12/7)) * color(blue)((1/x)) * color(green)((1/y^5))#

#color(white)("XXXX")##= 12/(7xy^5)#

Jul 9, 2015

Answer:

Simplify #24/14# to #12/7# by dividing by #2#. Apply the quotient rule and negative power rule of exponents.

Explanation:

#(24x^2y^2)/(14x^3y^7)#

Divide the numerator and denominator by #2#.

#(12x^2y^2)/(7x^3y^7)#

Apply the exponent quotient rule: #x^a/x^b=x^a-b#.

#(12x^(2-3)y^(2-7))/(7)# =

#(12x^(-1)y^(-5))/(7)# =

Apply the exponent negative power rule: #x^-a=1/x^a#.

#12/(7x^1y^5)# =

#12/(7xy^5)#