Question

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

Answer 1

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`

Answer 2

`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 separate`x^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))`