AngelaB97 one year ago can someone please explain this to me thoroughly because i keep getting confused on how o simplify this expression

1. AngelaB97

$(x^2/y^-2)^-2 (2y^-3/x^2)^3$

2. AngelaB97

|dw:1441669920633:dw|

3. AngelaB97

it is supposed to be 8/x^10y^13 bu i keep getting 8/x^10y^5

4. Jhannybean

$\large \left(\frac{x^2}{y^{-2}}\right)^{-2}\left(\frac{2y^{-3}}{x^2}\right)^3$Change the variables inside the parenthesis so they are all positive. $\large ( x^2y^2)^{-2}\left(\frac{2}{x^2y^3}\right)^3$Now that we've got all the variables INSIDE the parenthesis with positive powers, now we take care of the outer powers. Take the first one and put it over 1. $\frac{1}{(x^2y^2)^2}\cdot \left(\frac{2}{x^2y^3}\right)^3$

5. Jhannybean

Now let's distribute the powers through all variables inside the parenthesis. $\frac{1}{(x^2)^2(y^2)^2} \cdot \frac{2^3}{(x^2)^3(y^3)^3}$

6. AngelaB97

so since the exponent outside of the parentheses is negative aren't you supposed to flip it so it becomes y^2/x^2 ?

7. AngelaB97

r it that where i am making the mistake?

8. Jhannybean

Now let's distribute the powers through all variables inside the parenthesis. $\frac{1}{(x^2)^2(y^2)^2} \cdot \frac{2^3}{(x^2)^3(y^3)^3}$

9. AngelaB97

so since the exponent outside of the parentheses is negative aren't you supposed to flip it so it becomes y^2/x^2 ?

10. Jhannybean

Hmm... not exactly. We take care of each step individually.

11. AngelaB97

see, its (x^2/y^-2) isn't it supposed to be (y^2/x^2)^2

12. AngelaB97

@Jhannybean

13. Jhannybean

So we can write out $$\left(\dfrac{x^2}{y^{-2}}\right)^{-2}$$ as $$\dfrac{1}{\left( \dfrac{x^2}{\dfrac{1}{y^2}}\right)^2}$$

14. Jhannybean

Do you see whats happening here?

15. Jhannybean

I don't think it's proper to say $$\left(\dfrac{x^2}{y^{-2}}\right)^{-2} \iff \left(\dfrac{y^2}{x^2}\right)^2$$

16. AngelaB97

oh i understand now

17. AngelaB97

can you also help me with another problem please?

18. Jhannybean

Are you finished with this one?

19. AngelaB97

yes i am

20. Jhannybean

Alright :)

21. AngelaB97

okay so it's (2y^-1z/z^2)^-1 (y/3z^2)^2

22. Jhannybean

This is a new problem?

23. AngelaB97

yes

24. Jhannybean

$\left(\frac{2y^{-1}z}{z^2}\right)^{-1}\cdot \left(\frac{y}{3z^2}\right)^2$

25. Jhannybean

So like our previous problem, what will you want to do first?

26. AngelaB97

flip the problem?

27. Jhannybean

No, what do we do to all the negative variables within the parenthesis?

28. AngelaB97

we witch them dont we?

29. Jhannybean

By switch you mean... make them positive? :)

30. AngelaB97

yes

31. Jhannybean

You are correct. Therefore we will have $\left(\frac{2z}{yz^2}\right)^{-1}\cdot \left(\frac{y}{3z^2}\right)^2$

32. AngelaB97

yes so then it becomes (yz^2/2z)^1 ?

33. AngelaB97

multiplied by (y/3z^2)^2

34. AngelaB97

so first you make the numbers inside the parentheses positive, then you flip it whenever there's a negative exponent over the parentheses?

35. Jhannybean

$$\color{#0cbb34}{\text{Originally Posted by}}$$ @AngelaB97 so first you make the numbers inside the parentheses positive, then you flip it whenever there's a negative exponent over the parentheses? $$\color{#0cbb34}{\text{End of Quote}}$$ Im not understanding what you mean by that highlighted portion.

36. AngelaB97

as in whenever you have a negative exponent both outside and inside the parentheses, you have to deal with the numbers inside the parentheses?

37. Jhannybean

Yes, you deal with the numbers inside the ( ) first, making them + , then we take care of the outside ( ) by putting over 1. [i.e 1/ ( ) ]

38. Jhannybean

Does that make sense?

39. AngelaB97

yes exactly

40. Jhannybean

Cool! $$\checkmark$$ Glad you're following along.

41. Jhannybean

Now we have: $\left(\frac{2z}{yz^2}\right)^{-1}\cdot \left(\frac{y}{3z^2}\right)^2$ We're going to put the first one over 1 to make it +. $\frac{1}{\left(\dfrac{2z}{yz^2}\right)} \cdot \left(\frac{y}{3z^2}\right)^2$

42. Jhannybean

SOrry, bear with me, Im lagging a lot right now.

43. AngelaB97

don't worry you're helping me alot

44. Jhannybean

This becomes: $\frac{yz^2}{2z} \cdot \left(\frac{y}{3z^2}\right)^2$

45. Jhannybean

Now we can reduce our fraction. Remember, $$\dfrac{x^m}{x^n} = x^{m-n} \therefore \dfrac{z^2}{z} = z^{2-1} = z$$ Now we can rewrite this as: $\frac{yz}{2} \cdot \left(\frac{y}{3z^2}\right)^2$

46. AngelaB97

then yz/2 * y^2/9z^4

47. Jhannybean

exactly.

48. Jhannybean

We get $=\frac{y^3z}{18z^4}$ $\frac{z}{z^4} =z^{1-4} =z^{-3}$ $=\frac{y^3}{18z^{-3}} \iff \boxed{\frac{y^3z^3}{18}}$

49. Jhannybean

Whoop whoop wait a minute.

50. Jhannybean

Typo.

51. Jhannybean

we got $$z^{-3}$$ therefore it would be: $\frac{y^3z^{-3}}{18} \iff \boxed{\frac{y^3}{18z^3}}$

52. AngelaB97

gotcha, thanks sosososo much!!! :) <3

53. AngelaB97

you helped me out so much today

54. Jhannybean

No problem :) I've got to head off now. SO good luck on the rest!

55. Jhannybean