1. ## Could you please double-check my work

I'm working on a couple of problems here and I wanted to make sure my steps to my answers (in bold) are correct. The numbers in red are exponents.

1. Problem: 8 a-3
..................4a2

................= 2
..................a5

2. Problem:

(3x-2 y2)2
-12 x3 y-3

(3)2 (x-2)2 (y2)2
-12 x3 y-3

...3x4 y4
-12 x3 y-3

3x7
4y7

Thanks again.

2. Originally Posted by slykksta
I'm working on a couple of problems here and I wanted to make sure my steps to my answers (in bold) are correct. The numbers in red are exponents.

1. Problem: 8 a-3
..................4a2

................= 2
..................a5

2. Problem:

(3x-2 y2)2
-12 x3 y-3

(3)2 (x-2)2 (y2)2
-12 x3 y-3

...3x4 y4
-12 x3 y-3

3x7
4y7

Thanks again.
1. The work is fine.

(3)2 (x-2)2 (y2)2
-12 x3 y-3

In this, you simplified it to:

3x4 y4
-12 x3 y-3

Which is wrong, when you separate each component of the first part I copied, you end up getting the following on the numerator:

(3)^2 = 9

(x^-2)^2 = (x^-2)*(x^-2) = (x^(-2 + -2)) = (x^-4)

(y^2)^2 = (y^2)*(y^2) = y^4

The numerator is as such:

9x^-4*y^4

The whole fraction is:

9x^-4*y^4
-----------
-12x^3*y^-3

Now we have to use the law of inverses, which states that:

x^-1 = 1/x^1

So:

x^-4 = 1/x^4

And:

1/y^-3 = y^3

Now we have to rearrange the fraction:

9y^4*y^3
----------
-12x^3*x^4

Therefore, we have:

3y^7
-----
-4x^7

3. Hello, slykksta!

$\displaystyle 1)\;\;\frac{8a^{-3}}{4a^2}\;=\;\frac{2}{a^5}$ . . . . Right!

$\displaystyle 2)\;\;\frac{(3x^{-2}y^2)^2} {-12x^3y^{-3}}$

We have: .$\displaystyle \frac{3^2(x^{-2})^2(y^2)^2} {-12x^3y^{-3}} \;=\;\frac{9x^{-4}y^4}{-12x^3y^{-3}} \;=\;-\frac{3y^7}{4x^7}$