Question

How do you get rid of negative exponents in `\frac { 3x ^ { - 3} } { 2x ^ { 2} y ^ { - 2} \cdot y }`?

Answer 1

See a solution process belowL

Explanation:

First, use these rules for exponents to combine the `y` terms in the denominator:

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

`(3x^-3)/(2x^2y^-2 * y) => (3x^-3)/(2x^2y^color(red)(-2) * y^color(blue)(1)) => (3x^-3)/(2x^2y^(color(red)(-2)+color(blue)(1))) => (3x^-3)/(2x^2y^-1)`

Next, use this rule of exponents to eliminate the negative `x` exponent:

`x^color(red)(a)/x^color(blue)(b) = 1/x^(color(blue)(b)-color(red)(a))`

`(3x^color(red)(-3))/(2x^color(blue)(2)y^-1) => 3/(2x^(color(blue)(2)-color(red)(-3))y^-1) => 3/(2x^(color(blue)(2)+color(red)(3))y^-1) => 3/(2x^5y^-1)`

Now, use these rules of exponents to eliminate the negative `y` exponent:

`1/x^color(red)(a) = x^color(red)(-a)` and `a^color(red)(1) = a`

#3/(2x^5y^color(red)(-1)) => (3y^color(red)(- -1))/(2x^5) => (3y^color(red)(1))/(2x^5) => (3y)/(2x^5)#