## anonymous one year ago Combine as indicated by the signs: 4/y^2-9 + 5/y+3

1. UsukiDoll

do you know how to factor? and take the least common denominator?

2. anonymous

...unfortunatly i dont really I'm in the process of teaching myself.

3. UsukiDoll

$\frac{4}{y^2-9} + \frac{5}{y+3}$

4. UsukiDoll

there is a perfect square for the first fraction in the denominator ... a product of some integer with itself. $y^2-9$

5. UsukiDoll

in the form of $(a^2-b^2) =(a+b)(a-b)$ so if we let a = y^2 and b=9 we need to take the square root of y^2 and the square root of 9. Can you do that?

6. anonymous

wouldnt the square root of y^2 just be y and the square root of 9 is 3?

7. UsukiDoll

yay!

8. UsukiDoll

so since we have a = y and b = 3 we have $(y^2-9)=(y+3)(y-3)$

9. UsukiDoll

$\frac{4}{(y+3)(y-3)} + \frac{5}{y+3}$

10. UsukiDoll

now noticed how our denominators are different? :O we need to take the lcd.. since lcd is (y+3)(y-3) we need to multiply ?????????????? so the denominator in the second fraction is the same as the denominator in the first fraction

11. UsukiDoll

so what is missing in the denominator for the second fraction? I have y+3 but I don't have ?????

12. anonymous

y-3? im kinda lost.

13. UsukiDoll

you're on the right track

14. UsukiDoll

so we multiply y-3 on the numerator for the second fraction and multiply y-3 on the denominator for the second fraction as well

15. UsukiDoll

$\frac{4}{(y+3)(y-3)} + \frac{5}{y+3} \times \frac{y-3}{y-3}$ your fraction will look something like this

16. UsukiDoll

$\frac{4}{(y+3)(y-3)} + \frac{5(y-3)}{y+3(y-3)}$ so we distribute the 5 all over y-3

17. UsukiDoll

can you distribute the 5 towards y-3 ? it's like multiplying 5 times y and 5 times -3

18. anonymous

5y-15?

19. UsukiDoll

$\frac{4}{(y+3)(y-3)} + \frac{5y-15}{y+3(y-3)}$ yes

20. UsukiDoll

$\frac{4+5y-15}{(y+3)(y-3)}$ now we need to compute 4-15

21. anonymous

-11

22. UsukiDoll

correct

23. UsukiDoll

$\frac{5y-11}{(y+3)(y-3)}$

24. UsukiDoll

we can't factor anything out and there's nothing to cancel, so we're done

25. anonymous

ok i got it and can you look at one more problem and tell me why my answer is only partially right?

26. UsukiDoll

sure

27. anonymous

8-y/3y + y+2/9y - 2/6y i got 2y+23/9y

28. UsukiDoll

$\frac{8-y}{3y}+ \frac{y+2}{9y}-\frac{2}{6y}$ this?

29. anonymous

yes

30. UsukiDoll

ok so all of them have y in common but we notice that the denominators are different.. we also notice that the third fraction $\frac{2}{6}$ can be reduced . can you reduce that fraction for me?

31. anonymous

1/3

32. UsukiDoll

$\frac{8-y}{3y}+ \frac{y+2}{9y}-\frac{1}{3y}$ alright... now we noticed that the first and third fraction has the same denominator, so we can rewrite this fraction and solve . $\frac{8-y}{3y}-\frac{1}{3y}+ \frac{y+2}{9y}$ $\frac{8-y-1}{3y}+\frac{y+2}{9y}$ can you combine like terms on the first fraction ?

33. UsukiDoll

we just have to do this arithmetic 8-y-1 which can be rewritten as -y+8-1

34. anonymous

ok

35. anonymous

?

36. UsukiDoll

37. UsukiDoll

I can't combine variables... but I can combine numbers for -y+8-1

38. anonymous

so would it just be -1+8-1/9y?

39. UsukiDoll

no.. leave -y alone .. what is 8-1?

40. anonymous

7

41. UsukiDoll

$\frac{-y+7}{3y}+\frac{y+2}{9y}$ yes.. we still can't add these denominators.. we have the y's so we just need a number so 3 x ? = 9

42. anonymous

3

43. UsukiDoll

$\frac{3(-y+7)}{9y}+\frac{y+2}{9y}$

44. UsukiDoll

now distribute the 3 what is 3 times -y what is 3 x 7 ?

45. anonymous

3 x 7=21 3 x-y=-3y

46. UsukiDoll

$\frac{(-3y+21)}{9y}+\frac{y+2}{9y}$

47. UsukiDoll

so what is -3y+y what is 21+2

48. UsukiDoll

$\frac{(-3y+y+21+2)}{9y}$

49. anonymous

23 and -4y

50. UsukiDoll

23 is correct but for the y portion that's wrong

51. UsukiDoll

if you're adding a big negative number and a small positive number, your answer should go down.

52. UsukiDoll

$\frac{(-3y+y+23)}{9y}$

53. UsukiDoll

try again what is -3y+y

54. UsukiDoll

hmmm... let's ignore the y part ... let's just solve -3+1 think of it as... you want to buy something for $3 but you only have$1, how much more do you need to buy that product?

55. anonymous

2

56. UsukiDoll

yeah, but there's one problem... you're \$2 in the red in that example, so -2 is the answer

57. UsukiDoll

$\frac{(-2y+23)}{9y}$

58. UsukiDoll

you had the correct answer but you forgot the - sign on the 2y.. I highly suggest you practice on adding and subtracting with negative numbers

59. anonymous

i see now that makes more sense and i totally will work on it ha! thanks