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anonymous

  • one year ago

Combine as indicated by the signs: 4/y^2-9 + 5/y+3

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  1. UsukiDoll
    • one year ago
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    do you know how to factor? and take the least common denominator?

  2. anonymous
    • one year ago
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    ...unfortunatly i dont really I'm in the process of teaching myself.

  3. UsukiDoll
    • one year ago
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    \[\frac{4}{y^2-9} + \frac{5}{y+3}\]

  4. UsukiDoll
    • one year ago
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    there is a perfect square for the first fraction in the denominator ... a product of some integer with itself. \[y^2-9\]

  5. UsukiDoll
    • one year ago
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    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
    • one year ago
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    wouldnt the square root of y^2 just be y and the square root of 9 is 3?

  7. UsukiDoll
    • one year ago
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    yay!

  8. UsukiDoll
    • one year ago
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    so since we have a = y and b = 3 we have \[(y^2-9)=(y+3)(y-3) \]

  9. UsukiDoll
    • one year ago
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    \[\frac{4}{(y+3)(y-3)} + \frac{5}{y+3}\]

  10. UsukiDoll
    • one year ago
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    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
    • one year ago
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    so what is missing in the denominator for the second fraction? I have y+3 but I don't have ?????

  12. anonymous
    • one year ago
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    y-3? im kinda lost.

  13. UsukiDoll
    • one year ago
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    you're on the right track

  14. UsukiDoll
    • one year ago
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    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
    • one year ago
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    \[\frac{4}{(y+3)(y-3)} + \frac{5}{y+3} \times \frac{y-3}{y-3}\] your fraction will look something like this

  16. UsukiDoll
    • one year ago
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    \[\frac{4}{(y+3)(y-3)} + \frac{5(y-3)}{y+3(y-3)} \] so we distribute the 5 all over y-3

  17. UsukiDoll
    • one year ago
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    can you distribute the 5 towards y-3 ? it's like multiplying 5 times y and 5 times -3

  18. anonymous
    • one year ago
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    5y-15?

  19. UsukiDoll
    • one year ago
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    \[\frac{4}{(y+3)(y-3)} + \frac{5y-15}{y+3(y-3)}\] yes

  20. UsukiDoll
    • one year ago
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    \[\frac{4+5y-15}{(y+3)(y-3)} \] now we need to compute 4-15

  21. anonymous
    • one year ago
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    -11

  22. UsukiDoll
    • one year ago
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    correct

  23. UsukiDoll
    • one year ago
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    \[\frac{5y-11}{(y+3)(y-3)}\]

  24. UsukiDoll
    • one year ago
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    we can't factor anything out and there's nothing to cancel, so we're done

  25. anonymous
    • one year ago
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    ok i got it and can you look at one more problem and tell me why my answer is only partially right?

  26. UsukiDoll
    • one year ago
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    sure

  27. anonymous
    • one year ago
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    8-y/3y + y+2/9y - 2/6y i got 2y+23/9y

  28. UsukiDoll
    • one year ago
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    \[\frac{8-y}{3y}+ \frac{y+2}{9y}-\frac{2}{6y}\] this?

  29. anonymous
    • one year ago
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    yes

  30. UsukiDoll
    • one year ago
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    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
    • one year ago
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    1/3

  32. UsukiDoll
    • one year ago
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    \[\frac{8-y}{3y}+ \frac{y+2}{9y}-\frac{1}{3y}\] 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
    • one year ago
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    we just have to do this arithmetic 8-y-1 which can be rewritten as -y+8-1

  34. anonymous
    • one year ago
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    ok

  35. anonymous
    • one year ago
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    ?

  36. UsukiDoll
    • one year ago
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    so what is the answer?

  37. UsukiDoll
    • one year ago
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    I can't combine variables... but I can combine numbers for -y+8-1

  38. anonymous
    • one year ago
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    so would it just be -1+8-1/9y?

  39. UsukiDoll
    • one year ago
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    no.. leave -y alone .. what is 8-1?

  40. anonymous
    • one year ago
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    7

  41. UsukiDoll
    • one year ago
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    \[\frac{-y+7}{3y}+\frac{y+2}{9y}\] 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
    • one year ago
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    3

  43. UsukiDoll
    • one year ago
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    \[\frac{3(-y+7)}{9y}+\frac{y+2}{9y}\]

  44. UsukiDoll
    • one year ago
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    now distribute the 3 what is 3 times -y what is 3 x 7 ?

  45. anonymous
    • one year ago
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    3 x 7=21 3 x-y=-3y

  46. UsukiDoll
    • one year ago
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    \[\frac{(-3y+21)}{9y}+\frac{y+2}{9y}\]

  47. UsukiDoll
    • one year ago
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    so what is -3y+y what is 21+2

  48. UsukiDoll
    • one year ago
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    \[\frac{(-3y+y+21+2)}{9y}\]

  49. anonymous
    • one year ago
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    23 and -4y

  50. UsukiDoll
    • one year ago
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    23 is correct but for the y portion that's wrong

  51. UsukiDoll
    • one year ago
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    if you're adding a big negative number and a small positive number, your answer should go down.

  52. UsukiDoll
    • one year ago
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    \[\frac{(-3y+y+23)}{9y}\]

  53. UsukiDoll
    • one year ago
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    try again what is -3y+y

  54. UsukiDoll
    • one year ago
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    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
    • one year ago
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    2

  56. UsukiDoll
    • one year ago
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    yeah, but there's one problem... you're $2 in the red in that example, so -2 is the answer

  57. UsukiDoll
    • one year ago
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    \[\frac{(-2y+23)}{9y}\]

  58. UsukiDoll
    • one year ago
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    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
    • one year ago
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    i see now that makes more sense and i totally will work on it ha! thanks

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