Question

How do you simplify `\frac { 30x ^ { 8} y } { 6x ^ { 5} y ^ { 4} }`?

Answer 1

Use exponent laws and simplify any rational numbers. In this case, we get `=(5x^3)/(y^3)`.

Explanation:

All we do is use exponent laws.
=> Where if you divide an exponent with another of the same base, the exponents are subtracted.

`(30x^8y)/(6x^5y^4)`

So first off, we have two variables that are the same, `x` and `y`. We can simplify them but subtracting the exponents.

If we subtract from the top to bottom, the variable remains as the numerator.

Likewise, if we subtract from bottom to top, the variable remains as the denominator.

I want to avoid negative exponents to make it easier for myself, so I'll subtract the exponents in a way to avoid negatives.

`=(30x^3)/(6y^3)`

I did `8-5` and since I started from the top, the `x`-variable remains as a numerator

Same goes with `y`.

Now we simplify the fraction.

`=(5x^3)/(y^3)`

Hope this helps :)

Answer 2

`(5x^3)/y^3`

Explanation:

`color(blue)("This is what is happening using first principles")`

We are looking for values we can turn in to 1 as multiplying by 1 has no effect. Turning into one for multiply or divide is the same process as cancelling.

Split this up so that we have:

`(5xx6)/6xx(x^5xxx^3)/x^5 xxy/(yxxy^3)`

`5xx6/6xxx^5/x^5xx x^3xx y/yxx1/y^3`

`5xx1xx1xx x^3xx1xx1/y^3`

`(5x^3)/y^3`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("The same process but using the shortcuts of cancelling out")`

`(30x^8y)/(6x^5y^4)" "->" "(cancel(30)^5color(white)(..)x^(cancel(color(white)(.)8)^3)color(white)(...)cancel(y))/(cancel(6)^1cancel(x^5)color(white)(.)y^(cancel(color(white)(.)4)^3))" "->" "(5x^3)/(y^3)`