Another one - checking son's homework

[dbush]

Here's another problem I would like for you to review my answer...I can't seem to get the problem to show up correctly. The x²y on the bottom is supposed to be the bottom half of (4xy²)^-1.

Simplify:

8x²y^-3 X (4xy²)^-1
x^-2y x²y


Drum roll please....my answer is
8x⁵
4y⁵

My brain is fried so I'm not really sure at this point how I got here, but it made sense when I was working the problem.
Thanks for your help!

 

[tkhunny]

Perhaps you should have boiled, rather than fried?

Collect constants: 8 * 4^(-1) = 8/4 = 2

Collect x's: (x^2)(x)(x^(-1))/[(x^(-2))(x^2)] = x^2 -- or is the big "X" a multiplication symbol?

You do the y's

 

[dbush]

The big "X" is a multiplication symbol. Sorry!!

 

[Deleted member 4993]

To avoid confusion, start using "*" (as in m*n) to indicate multiplication.

 

[dbush]

Yes, I can see where the * would be clearer. I appreciate the advice. I would be grateful if anyone can make sense of this problem and check my answer. Many thanks!

 

[pappus]

Here's another problem I would like for you to review my answer...I can't seem to get the problem to show up correctly. The x²y on the bottom is supposed to be the bottom half of (4xy²)^-1.

Simplify:

8x²y^-3 X (4xy²)^-1
x^-2y x²y

...

1. I assume that you know the basic laws to calculate powers(?), for instance \(\displaystyle \displaystyle{\frac1{x^{-2}} = x^2}\) or \(\displaystyle \displaystyle{x^a \cdot x^b = x^{a+b}}\)

2. If so re-write your term as a product:

\(\displaystyle \displaystyle{\frac{8x^2 y^{-3}}{x^{-2}y} \cdot \frac{(4xy^2)^{-1}}{x^2y} = \frac84 \cdot x^{2} \cdot x^{-1}\cdot x^{2}\cdot x^{-2} \cdot y^{-3}\cdot y^{-2}\cdot y^{-1}\cdot y^{-1}}\)

3. Collect the exponents:

\(\displaystyle \displaystyle{2x^{2-1+2-2} \cdot y^{-3-2-1-1} = \frac{2x}{y^7}}\)

 

[srmichael]

Here's another problem I would like for you to review my answer...I can't seem to get the problem to show up correctly. The x²y on the bottom is supposed to be the bottom half of (4xy²)^-1.

Simplify:

8x²y^-3 X (4xy²)^-1
x^-2y x²y


Drum roll please....my answer is
8x⁵
4y⁵

My brain is fried so I'm not really sure at this point how I got here, but it made sense when I was working the problem.
Thanks for your help!

So I think you have:

\(\displaystyle \displaystyle \frac{8x^2y^{-3}}{x^{-2}y}\cdot\ \frac{(4xy^2)^{-1}}{x^2y}\)

which can be rewritten as:

\(\displaystyle \displaystyle \frac{(8x^2y^{-3})(4xy^2)^{-1}}{(x^{-2}y)(x^2y)}\)

There are a few different ways you can approach these kinds of problems. Here is one way:

\(\displaystyle \displaystyle \frac{(8x^2y^{-3})(4^{-1}x^{-1}y^{-2})}{(x^{-2}y)(x^2y)}\)

\(\displaystyle \displaystyle \frac{8\cdot4^{-1}x^1y^{-5}}{x^{0}y^2}\)

\(\displaystyle \displaystyle \frac{8x}{4y^5y^2}\)

\(\displaystyle \displaystyle \frac{2x}{y^7}\)

 

[dbush]

Yes, I see my mistake now. Thank you! This greatly helps me explain the problem to my son who is struggling in algebra. Struggling is an understatement.