How can you solve (6x^-2y^7)/(3x^-7y^-3)?

2 Answers
Mar 28, 2018

See a solution process below:

Explanation:

First, rewrite the expression as:

6/3(x^-2/x^-7)(y^7/y^-3) =>

2(x^-2/x^-7)(y^7/y^-3)

Now, use this rule of exponents to simplify the expression:

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

2(x^color(red)(-2)/x^color(blue)(-7))(y^color(red)(7)/y^color(blue)(-3)) =>

2x^(color(red)(-2)-color(blue)(-7))y^(color(red)(7)-color(blue)(-3)) =>

2x^(color(red)(-2)+color(blue)(7))y^(color(red)(7)+color(blue)(3)) =>

2x^5y^10

Mar 28, 2018

2 x^5 y^10

Explanation:

any thing with a negative power will switch its place up and down. so, (6 x^7y^7 y^3)/(3x^2)
now as on the numerator we have y^7 and y^3 adds up = y^10 (a^b * a^c =a^(c+b))
(6x^7y^10)/(3x^2)
we can separatex^7 = x^2 and x^5
now solve it: (cancel(6)color(red)(2)cancel(x^2)x^5 y^10)/ (cancel(3)cancel(x^2))