Question

How do you simplify `\frac { 3y ^ { 4} } { 6y ^ { - 4} }`?

Answer 1

`(3y^4)/(6y^(-4))=y^8/2`

Explanation:

Before we solve the question, let us consider `a^5xxa^3` and `a^5/a^3`.

`a^5xxa^3` is equal to `(axxaxxaxxaxxa)xx(axxaxxa)`, which is nothing but `axxaxxaxxaxxaxxaxxaxxa=a^8` i.e. `a^(5+3)`.

What about `a^5/a^3`? Well this is `(axxaxxaxxaxxa)/(axxaxxa)` or `axxa`, as three `a's` in numerator and denominator cancel out and only two `a's` in numerator are left i.e.

`(axxaxxaxxaxxa)/(axxaxxa)=a^2` or we may write it as `a^(5-3)`, meaning thereby that `1/(axxaxxa)=a^(-3)`, which is nothing but `1/a^3`.

What about `(3y^4)/(6y^(-4))`?

Using above we can write it as `(3y^4)/(6xx1/y^4)`.

But dividing by `1/y^4` is equivalent to multiplying by its reciprocal or multiplicative inverse (i.e. interchanging numerator and denominator) `y^4/1` or `y^4`.

Hence `(3y^4)/(6y^(-4))=(3y^4)/(6xx1/y^4)=(3y^4)/6xxy^4/1`

= `(cancel3y^4)/(2cancel6)xxy^4/1`

= `y^(4+4)/2`

= `y^8/2`

Answer 2

`y^8/2`

Explanation:

`color(blue)("Concept used")`

By example
`x xx x = x^1 xx x^1 = x^(1+1) = x^2`

`x^1=x" but "x^(-1) = 1/x`
`x^2=x xx x" but "x^(-2)=1/x^2 = 1/(x xx x)`
..........................................................
`1/x^(1) = 1/x" but " 1/(x^(-1)) = 1xx x^1 =x`

`1/(x^2) = 1/(x xx x)" but "1/(x^(-2))=1xx x^2 = x^2`

Just to make this absolutely clear:

`2/x^(-3) = 2xxx^3 = 2x^3`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Using this method")`

Given:`" "(3y^4)/(6y^(-4)`

Rather than jump straight in lets do this in parts.

Write as `" "3xx1/6xxy^4xx1/(y^(-4))`

`color(brown)("Dealing with the numbers")`

`3xx1/6=3/6`

But 3 will divide into 3 and 6 exactly so divide top and bottom by 3

`color(brown)((3-:3)/(6-:3) = 1/2)`

`color(brown)("................................................................................")`
`color(brown)("Dealing with the y's")`

As in the example `1/y^(-4) = 1xxy^4 = y^4`

So `color(brown)(" "y^4xx1/(y^(-4))" " =" " y^4xxy^4" "=" "y^(4+4)" " =" " y^8)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Putting it all together")`

`color(blue)((3y^4)/(6y^(-4))" " =" " y^8/2)`