# Thread: [SOLVED] Fraction using exponents and parenthesis

1. ## [SOLVED] Fraction using exponents and parenthesis

Hi everyone first time here, got 2 questions, I'm pretty sure I got the first one rite but would like a second opinion and would like to no how to solve the second one....

First one is:

Simplify:

And would I be rite in saying the answer is :

Second is this:

I'm confused as to how I should handle the exponents which are both inside and outside of the parenthesis ?? Any help will be greatly appreciated

2. Originally Posted by JamesC1705
Hi everyone first time here, got 2 questions, I'm pretty sure I got the first one rite but would like a second opinion and would like to no how to solve the second one....

First one is:

Simplify:

And would I be rite in saying the answer is :

Second is this:

I'm confused as to how I should handle the exponents which are both inside and outside of the parenthesis ?? Any help will be greatly appreciated
The first one is good. You can simplify it further:

$\displaystyle \frac{9x^2}{12y^2} = \frac{3x^2}{4y^2}$

For the second one:

$\displaystyle \frac{(4x^{3}y)^2}{x^5} = \frac{4x^{3}y \times 4x^{3}y}{x^5}$

OR

$\displaystyle \frac{(4x^{3}y)^2}{x^5} = \frac{(4^2)(x^3)^{2}(y^2)}{x^5}$

Can you do it now?

3. ## getting there ....

First up the code option is rather exciting - gonna have a go at that is a sec...

With the first problem am I rite in saying you have divided by 3 because it's the greatest common factor?

Now... I'm going to try coding my answer to the second problem:

Regarding $\displaystyle \frac{(4x^{3}y)^2}{x^5} = \frac{4x^{3}y \times 4x^{3}y}{x^5}$

So effectively we are performing $\displaystyle \frac{4x^{3}y}{x^5}$ twice to acount for the fact that the top part is raised to the power of 2?

So $\displaystyle \frac{4x^{3}y}{x^5} = \frac{4y}{x^2}$

Meaning $\displaystyle \frac{4x^{3}y \times 4x^{3}y}{x^5} = \frac{16y^2}{x^4}$ ??

4. Originally Posted by JamesC1705

First up the code option is rather exciting - gonna have a go at that is a sec...

With the first problem am I rite in saying you have divided by 3 because it's the greatest common factor?

Now... I'm going to try coding my answer to the second problem:

Regarding $\displaystyle \frac{(4x^{3}y)^2}{x^5} = \frac{4x^{3}y \times 4x^{3}y}{x^5}$

So effectively we are performing $\displaystyle \frac{4x^{3}y}{x^5}$ twice to acount for the fact that the top part is raised to the power of 2?

So $\displaystyle \frac{4x^{3}y}{x^5} = \frac{4y}{x^2}$

Meaning $\displaystyle \frac{4x^{3}y \times 4x^{3}y}{x^5} = \frac{16y^2}{x^4}$ ??

For your first question. Yes. 3 is a factor of both 9 and 12, so you can divide both by 3.
--------------------------------------

$\displaystyle \frac{4x^{3}y \times 4x^{3}y}{x^5} = \frac{16 x^{6}y^2}{x^5}$

Now reduce x

5. Originally Posted by harish21
No.

$\displaystyle \frac{4x^{3}y \times 4x^{3}y}{x^5} = \frac{16 x^{6}y^2}{x^5}$

Now reduce x
Okkaaayy so I get $\displaystyle \frac{16 xy^2}{Nothing?}$ or is the answer just $\displaystyle 16xy^2$ ??

6. Originally Posted by JamesC1705
Okkaaayy so I get $\displaystyle \frac{16 xy^2}{Nothing?}$ or is the answer just $\displaystyle 16xy^2$ ??
yes. but dont write nothing in the denominator!

you get: $\displaystyle \frac{16 xy^2}{1} = 16xy^2$

7. Originally Posted by harish21
yes. but dont write nothing in the denominator!

you get: $\displaystyle \frac{16 xy^2}{1} = 16xy^2$
Yea I no I was joking - I was never gonna write nothing haha I thought it would be one... phew ok so tyvm just another 12 questions to go!

Thanks for your help - James