## Destinyyyy one year ago Cant remember how to solve this...

1. anonymous

Multiply. 5xy^5 / 3x^5y^3 * 6x^5y^6 / 25x^7y^10

2. anonymous

-6x^5y^6 **

3. phi

y^5 means y*y*y*y*y if you divide that by y^3 $\frac{y\cdot y\cdot y\cdot y\cdot y}{y\cdot y\cdot y} = y\cdot y \cdot \frac{ y\cdot y\cdot y}{y\cdot y\cdot y}= y\cdot y= y^2$ or the fast way: if you are dividing , subtract top exponent minus bottom exponent i.e. $\frac{y^5}{y^3}= y^2$ notice 5-3=2

4. anonymous

Okay thats ringing a bell...

5. anonymous

but dont I need to simplify 25 and -6 into two smaller numbers?

6. phi

is this the original problem: $\frac{5xy^5}{3x^5y^3} \cdot \frac{-6x^5y^6 }{25x^7y^{10}}$

7. anonymous

Yes

8. phi

if so, we multiply top times top and bottom times bottom $\frac{5xy^5\cdot-6x^5y^6}{3x^5y^325x^7y^{10}}$ and you can reorder the top (and bottom) I would just start dividing the bottom into the top as much as possible

9. phi

for example, I already figured out y^5/y^3 = y^2 so we can simplify to $\frac{5xy^5\cdot-6x^5y^6}{3x^5y^325x^7y^{10}} \\ \frac{5xy^2\cdot-6x^5y^6}{3x^5\cdot 25x^7y^{10}}$

10. phi

3 in the bottom can divide into -6 up top, to get -2 I would move the minus sign "out front". we get $-\frac{5xy^2\cdot2x^5y^6}{x^5 \cdot 25x^7y^{10}}$

11. phi

notice you have x^5 in the top and in the bottom. they "cancel out" (anything divided by itself is 1) so you have $-\frac{5xy^2\cdot2x^5y^6}{x^5 \cdot 25x^7y^{10}} = -\frac{5xy^2\cdot2y^6}{25x^7y^{10}}$

12. phi

notice you have 5 up top and 25 down below 25 is 5*5 so we have a 5/5 that cancel, and 5 remains down below.

13. anonymous

Im trying to follow but im getting lost... I remember having to do the first step as --- 5*x*y^5/ 3*x^5*y^3 * 2*3*x^5*y^6/ 5*5*x^7*y^10

14. anonymous

And then minus the exponents and all.

15. anonymous

So I now have - 2/5

16. phi

the way I do these (where everything up top is multiplied and everything down below is multiplied) is ignore everything except one thing up top and the same thing down below for example, if we look just at 5 up top and 25 below $\frac{5}{25} = \frac{5}{5\cdot 5} = \frac{1}{5}$ and I would replace the 5 up top by 1 and the 25 below with 5 and *everything else stays the same*

17. phi

*** 5*x*y^5/ 3*x^5*y^3 * 2*3*x^5*y^6/ 5*5*x^7*y^10 *** yes that is good. (but is there supposed to be a minus sign in there) ?

18. anonymous

Um okay?

19. anonymous

Yes because of -6

20. phi

ok. so if we just at the numbers we have $- \frac{5\cdot 2 \cdot 3}{3 \cdot 5 \cdot 5 }$ one pair of 5's cancel, and so does the pair of 3's so yes we get $- \frac{2}{5}$

21. phi

now do the x's $\frac{x\cdot x^5}{x^5\cdot x^7}$ we could make the top x^6 and the bottom x^12 but I try to divide first. for example x^5 divided by x^5 cancels (i.e.=1)

22. phi

is it clear how to simplify the x's or is it confusing?

23. anonymous

So your talking about combining then but to try to divide first..

24. anonymous

Do you mean subtract??

25. phi

the rules might be confusing, but remember this: for example x^5 means x*x*x*x*x and if we divide by x^7 = x*x*x*x*x*x*x i.e. $\frac{x\cdot x\cdot x\cdot x\cdot x}{ x\cdot x\cdot x\cdot x\cdot x\cdot x\cdot x}$

26. anonymous

I have this-- - 2y^2x^-2/ 5x^-4 y^-4

27. phi

every time you have an "x" up top and "x" down below, they cancel out so just cancel them out in pairs.

28. anonymous

29. phi

what do you get for $\frac{x\cdot x^5}{x^5\cdot x^7}$?

30. phi

I would try to keep the exponents positive (less confusing) $\frac{x\cdot \cancel{x^5}}{\cancel{x^5}\cdot x^7} = \frac{x}{ x^7}= \frac{1}{x^6}$

31. anonymous

Thank you for taking away 40 minutes of my time that I needed.

32. anonymous

This question should of taken less then 5 minutes to solve.

33. anonymous

Thank you for explaining things. But I did not need that. All I needed was a refresher. Im studying for my final exam.

34. phi

oops $-\frac{2}{5} \frac{1}{x^6y^2}$

35. anonymous

I figured that out 6 minutes ago..