How do you simplify #(2x^-3y)/( 10y^-2)#?

1 Answer
Sheffy
May 6, 2016

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

Explanation:

We know that #x^-n=1/x^n#

Applying this property to the variables:

Method 1:

#(2xx(1/x^3)xxy)/(2xx5xx(1/y^2))#

#=y/[(5xxx^3)xx1/y^2]#

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

Method 2:

We can write all the variables in the numerator as

#[2xxx^-3 xx y xxy^-(-2)]/10#

#=[x^-3 xx y xxy^2]/5# #color(red)("cancelling 2")#

#=[x^-3 xx y^3]/5#

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

Related questions
See all questions in Negative Exponents
Impact of this question
1551 views around the world
You can reuse this answer
Creative Commons License