Question

How do you simplify `(\frac { 44x ^ { 3} y ^ { - 4} z ^ { 2} } { - 4x ^ { 4} y ^ { 5} z ^ { - 8} } )` ?

Answer 1

See a solution process below:

Explanation:

First, rewrite the expression as:

`(44/-4)(x^3/x^4)(y^-4/y^5)(z^2/z^-8) =>`

`-11(x^3/x^4)(y^-4/y^5)(z^2/z^-8)`

Next, use this rule for exponents to simplify the `x` and `y` terms:

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

`-11(x^color(red)(3)/x^color(blue)(4))(y^color(red)(-4)/y^color(blue)(5))(z^2/z^-8) =>`

`-11(1/x^(color(blue)(4)-color(red)(3)))(1/y^(color(blue)(5)-color(red)(-4)))(z^2/z^-8) =>`

`-11(1/x^(color(blue)(4)-color(red)(3)))(1/y^(color(blue)(5)+color(red)(4)))(z^2/z^-8) =>`

`-11(1/x^1)(1/y^9)(z^2/z^-8) =>`

`-11/(x^1y^9)(z^2/z^-8)`

Then, use this rule for exponents to simplify the `z` term:

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

`-11/(x^1y^9)(z^color(red)(2)/z^color(blue)(-8)) =>`

`-11/(x^1y^9)(z^(color(red)(2)-color(blue)(-8))) =>`

`-11/(x^1y^9)(z^(color(red)(2)+color(blue)(8))) =>`

`-11/(x^1y^9)(z^10) =>`

`-(11z^10)/(x^1y^9)`

Now, use this rule for exponents to complete the simplification of the `x` term:

`a^color(red)(1) = a`

`-(11z^10)/(x^color(red)(1)y^9) =>`

`-(11z^10)/(xy^9)`