1. anonymous

2. anonymous

it's all about making the the bottom fraction 1... just a sec

3. anonymous

ya i get that but i confuse myself... i know you need a common demnominator..

4. anonymous

ugh i keep messing uppppp

5. anonymous

for yours, you have to get simple fraction over simple fraction first...

6. anonymous

$\frac{ 2-\frac{ 4 }{ x+2 } }{5+\frac{ 10 }{ x+2 } }$ is what you have but you need a simple fraction over a simple fraction. to get this, we start with a common denominator for the top expression... $2-\frac{ 4 }{ x+2 }=2\left( \frac{ x+2 }{ x+2 } \right)-\frac{ 4 }{ x+2 }=\frac{ 2x+4 }{ x+2 }-\frac{ 4 }{ x+2 }=\frac{ 2x }{ x+2 }$ does this make sense?

7. anonymous

yes that makes sense

8. anonymous

so you do the same to the bottom fraction?

9. anonymous

so now we do the same with the bottom expression. Would you like to try?

10. anonymous

yes give me a min to do it

11. anonymous

okay

12. anonymous

|dw:1440524508903:dw|

13. anonymous

excuse the terrible penmanship lol

14. anonymous

very good!!!

15. anonymous

so now we have $\Large{\frac{ \frac{ 2x }{ x+2 } }{ \frac{ 5x+20 }{ x+2 } }}$ right?

16. anonymous

|dw:1440524931675:dw|

17. anonymous

and then i do this?

18. anonymous

yes, do you know why?

19. anonymous

because its like dividing regular fractions... you flip and multiply... sooo the (x+2) crosses out right?

20. anonymous

well, the idea is to make an equivalent expression that is a simple fraction rather than a complex fraction. This is because we humans have a much easier time understanding what a simple fraction represents compared to a complex fraction. to make a simple fraction from a complex fraction, we simply manipulate the bottom fraction so that it is 1. we do this by multplying by the reciprocal of the bottom fraction, like this... $\Large{\frac{ \frac{ 2x }{ x+2 } }{ \frac{ 5x+20 }{x+2 } }=\frac{ \frac{ 2x }{ x+2 }\cdot \left( \frac{ x+2 }{ 5x+20 } \right) }{ \frac{ 5x+20 }{x+2 }\cdot \left( \frac{ x+2 }{ 5x+20 } \right)}=\frac{ \frac{ 2x }{ x+2 }\cdot \left( \frac{ x+2 }{ 5x+20 } \right) }{ \normalsize{1}}}$

21. anonymous

anything over 1 is just itself so that's why we end up with $\frac{ 2x }{ x+2 }\cdot \frac{ x+2 }{ 5x+20 }$ and we simplify from there

22. anonymous

does that make sense?

23. anonymous

yes so you factor 5x+20... to get 5(x+4)

24. anonymous

$\frac{ 2x }{ 5(x+4)}$ is the answer

25. anonymous

or should i not factor it?

26. anonymous

oh waiiittt hold on

27. anonymous

wait nvm thats my final answer lol

28. anonymous

to factor or not to factor is a matter of preference... you still have reduced to lowest terms.

29. anonymous

but does the process make sense to you?

30. anonymous

yeah it makes sense to focus on the numerator and denominator separately before you put it all together.. i was trying to do it all in one stepp. so thats my final correct answer?

31. anonymous

yep

32. anonymous

good job!

33. anonymous

thank you for actually teaching me.. no one else takes the time to do that i have 5 more to do lol. but would you be willing to check one of them for me after i do it.

34. anonymous

sure

35. anonymous

36. anonymous

and i think that can be reduced?

37. anonymous

correct, it can be reduced

38. anonymous

well this is a silly brain fart but i would reduce all the terms by 2?

39. anonymous

$\frac{ 4x-12 }{ 5x}$

40. anonymous

yep, that's the extent of it

41. anonymous

one more question lol

42. anonymous

43. anonymous

for the first question the denominator isnt a fraction...

44. anonymous

you can make it one... $5+a=\frac{ 5+a }{ 1 }$

45. anonymous

ohhhh wait i got thiis

46. anonymous

yes except it is $$25-a^2 \text{ not } a^2 - 25$$

47. anonymous

gotchaaa.. but does the order i write it affect it? i thought it was a difference of squares either way

48. anonymous

true, but $$25-a^2 \ne a^2-25$$ unless $$a=\pm 5$$

49. anonymous

ahh i gotchaa. seriously thank you so much! youre amazingg! i might message you when im in need of a lesson lol

50. anonymous

right on! good luck, have a great day and keep on learning!