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pooja195

  • one year ago

@mathmate

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  1. pooja195
    • one year ago
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    @mathmate

  2. pooja195
    • one year ago
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    factor 25-y^2

  3. pooja195
    • one year ago
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    I would just square root both right? :P

  4. mathmate
    • one year ago
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    difference of two squares!

  5. pooja195
    • one year ago
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    (x+5)(x-5) ?

  6. mathmate
    • one year ago
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    Yep!

  7. pooja195
    • one year ago
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    Thanks :)

  8. mathmate
    • one year ago
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    Now, Simplify \(\Large \frac{3x^3}{6x^2}\)

  9. pooja195
    • one year ago
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    \[\frac{ 3*x*x*x }{ 3*2*x*x }\] \[\frac{ x }{ 2 }\]

  10. mathmate
    • one year ago
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    What you should do \(whenever\) you divide out anything from the denominator is to make sure the factor cannot be zero. 3 is not zero, so nothing needs to be done! But...

  11. pooja195
    • one year ago
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    -_-

  12. mathmate
    • one year ago
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    You would specify that the answer is \(\frac{x}{2}\) for x\(\ne\)0

  13. mathmate
    • one year ago
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    is that ok?

  14. mathmate
    • one year ago
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    Remember whenever we cancel a factor from the denominator, we are not allowed to cancel unless the factor is \(not\) equal to zero.

  15. mathmate
    • one year ago
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    Most teachers will take half the points away if you forgot that, or some of these conditions.

  16. pooja195
    • one year ago
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    O_O

  17. mathmate
    • one year ago
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    That was over copied! Simplify \(\Large \frac{x(x^2+6)}{x^2}\)

  18. pooja195
    • one year ago
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    (x^2+6) / x

  19. mathmate
    • one year ago
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    Condition?

  20. pooja195
    • one year ago
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    \[x \neq1\]

  21. mathmate
    • one year ago
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    You cancelled x, so the condition is x\(\ne\)0

  22. mathmate
    • one year ago
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    It's what you cancelled should not equal to zero.

  23. mathmate
    • one year ago
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    x\(\ne\)1 is for cancelling (x-1)

  24. mathmate
    • one year ago
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    are we good?

  25. pooja195
    • one year ago
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    Yes

  26. mathmate
    • one year ago
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    ok, ready for the next one!

  27. pooja195
    • one year ago
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    mhm

  28. mathmate
    • one year ago
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    Simplify \(\Large \frac{p^2-2p+1}{p^2-1}\)

  29. mathmate
    • one year ago
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    Remember to specify the conditions if you cancelled any factor.

  30. pooja195
    • one year ago
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    \[\frac{ (p−1)(p−1) }{ p^2-1}\]

  31. mathmate
    • one year ago
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    Continue! There are factors for the denominator.

  32. mathmate
    • one year ago
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    diff of two squares!

  33. mathmate
    • one year ago
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    In the worst case, use the quadratic formula!

  34. pooja195
    • one year ago
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    \[\frac{ (p-1)(p-1) }{ (p+1)(p-1) }\]

  35. mathmate
    • one year ago
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    Excellent, you can now finish off. Tag on the conditions whenever you cancel!

  36. pooja195
    • one year ago
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    \[\frac{ p-1 }{ p+1 }~~~~~~~ x \neq 1\]

  37. pooja195
    • one year ago
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    ** -1

  38. mathmate
    • one year ago
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    x\(\ne\)1 is correct, because it is the same as (x-1)\(\ne\)0

  39. mathmate
    • one year ago
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    is that clear how we got x\(\ne\)1 ?

  40. pooja195
    • one year ago
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    yes

  41. mathmate
    • one year ago
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    ready for another?

  42. pooja195
    • one year ago
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    no more of these ;-;

  43. mathmate
    • one year ago
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    This is going to be all of these for the rest of 11! Just one more of these simple ones. Once you can do the simple ones, you'll even like the more complicated ones! lol Simplify \(\Large \frac{3(4-m)}{6(m-4)}\)

  44. pooja195
    • one year ago
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    2

  45. mathmate
    • one year ago
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    you get the idea, but a little too fast ! :(

  46. mathmate
    • one year ago
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    Remember m-4 = -(4-m), and 3/6=1/2

  47. mathmate
    • one year ago
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    ... and the conditions!

  48. pooja195
    • one year ago
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    x /= 1/3 , 3/6

  49. mathmate
    • one year ago
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    Well, we have \(\Large \frac{3(4-m)}{6(m-4)}\) = \(\Large \frac{-3(m-4)}{6(m-4)}\) =\(\Large \frac{-3(m-4)}{3*2(m-4)}\) so we cancel 3 and (m-4) but we know that 3\(\ne\)0, so we don't have to specify as a condition, the other one is m-4\(\ne\)0. If we add 4 to each side, we have m\(\ne\)4

  50. mathmate
    • one year ago
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    So the final answer is -1/2, where m\(\ne\)4

  51. mathmate
    • one year ago
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    are we good, especially with conditions? We'll be working with conditions and factorization throughout ch. 11

  52. pooja195
    • one year ago
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    yessss

  53. mathmate
    • one year ago
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    34. simplify \(\Large \frac{x^2-9}{x^2-5x-6}\)

  54. pooja195
    • one year ago
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    \[\frac{ (x+3)(x-3) }{ (x-6)(x+1)}\]

  55. mathmate
    • one year ago
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    anything else to do or to write?

  56. mathmate
    • one year ago
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    (note: the factoring is correct! Well-done!)

  57. pooja195
    • one year ago
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    nope thats all : P

  58. mathmate
    • one year ago
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    If you did not cancel anything, then you make a note: the expression is already in its simplest form. (and no conditions if nothing cancelled)

  59. pooja195
    • one year ago
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    ok :)

  60. mathmate
    • one year ago
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    Simplify if possible \(\Large \frac{3x-5}{25-30x+9x^2}\)

  61. pooja195
    • one year ago
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    i need help ;-;

  62. mathmate
    • one year ago
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    Remember steps for factoring: 1. take out GCF (common factors) 2. Check if it is difference of 2 squares (Must be a difference, and only has two terms) 3. Check if it is a perfect square: First and last terms are perfect squares, and middle term is twice the product of the square-roots of the two end terms. Example: a^2+2ab+b^2 : End terms are perfect squares, middle = 2(a)(b), so yes, this is a perfect square, equal to (a+b)^2. 4. Try factoring.

  63. mathmate
    • one year ago
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    1. any common factors (in the denominator)?

  64. pooja195
    • one year ago
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    no

  65. mathmate
    • one year ago
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    2. Is it diff. of 2 squares?

  66. pooja195
    • one year ago
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    no

  67. mathmate
    • one year ago
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    3. Is it a perfect square?

  68. pooja195
    • one year ago
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    no

  69. mathmate
    • one year ago
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    Hmm, let's see. Square-root of first term = sqrt(25) =5 square-root of last term =sqrt(9x^2)= 3x (so far so good, both terms are perfect squares)

  70. mathmate
    • one year ago
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    Now do a foil on (5-3x) and see if the middle term fits. (5-3x)(5-3x)=5^2-15x-15x+9x^2=5^2-30x+9x^2 Yay, so it's a perfect, equal to (5-3x)^2

  71. mathmate
    • one year ago
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    So now can you complete it?

  72. mathmate
    • one year ago
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    Shall we go back to perfect squares?

  73. pooja195
    • one year ago
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    \[\frac{ 1 }{ 3x-5 }\]

  74. mathmate
    • one year ago
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    Watch out: 5-3x = -1 (3x-5) and also since we cancelled 3x-5, we specify the condition x\(\ne\)5/3

  75. pooja195
    • one year ago
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    >.<

  76. mathmate
    • one year ago
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    So the answer is:

  77. pooja195
    • one year ago
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    1/3x-5 x/= 5/3

  78. mathmate
    • one year ago
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    Yep, that's correct!

  79. mathmate
    • one year ago
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    For perfect squares, it has to satisfy 3 conditions 1. end terms are perfect squares, i.e. you can find the square roots without leaving the radical. 2. middle term equals twice the square-roots of the end terms. 3. the middle term has the same sign of the factored terms.

  80. mathmate
    • one year ago
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    We check that with an example Factor 81x^2- xy +64y^2.

  81. mathmate
    • one year ago
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    Note that condition one means that both end terms must be positive. Here first term = 81x^2, square root = 9x

  82. mathmate
    • one year ago
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    Last term = 64 y^2, square-root = ?

  83. mathmate
    • one year ago
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    @pooja195 What is the square-root of the last term?

  84. pooja195
    • one year ago
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    3 :/ ?

  85. mathmate
    • one year ago
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    Last term = 64 y^2, square-root = ?

  86. pooja195
    • one year ago
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    8y

  87. mathmate
    • one year ago
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    Good. What's twice the product of the square-roots?

  88. pooja195
    • one year ago
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    huh ?

  89. mathmate
    • one year ago
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    We need to multiply together the square-roots (of the first and third terms), and double it.

  90. pooja195
    • one year ago
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    512? :/

  91. pooja195
    • one year ago
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    im so confused :/

  92. mathmate
    • one year ago
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    the square roots are 9x and 8y,so the product is 72xy, and doubling it shoud give 144xy. This should match the middle term (if the expression is a perfect square). It doesn't because I forgot to fill in the 144 in the middle. So we conclude that (9x+8y)^2=81x^2+144xy+64y^2=(9x+8y)^2.

  93. mathmate
    • one year ago
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    |dw:1433718072207:dw|

  94. mathmate
    • one year ago
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    |dw:1433718524311:dw|

  95. mathmate
    • one year ago
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    So if we are given 81x^2+144xy+64y^2 We can do the reverse of FOIL, assuming that it is a perfect square. We find 9x by sqrt, and 8y by sqrt. Then we test if 2(9x)(8y)=144xy, if it's equal (9x+8y)^2 are the factors.

  96. mathmate
    • one year ago
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    |dw:1433718696561:dw|

  97. pooja195
    • one year ago
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    Thats so much more better :)

  98. pooja195
    • one year ago
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    I understand this a bit more :o

  99. mathmate
    • one year ago
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    Good! An image is worth 1000 words, I was told!

  100. pooja195
    • one year ago
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    xD

  101. mathmate
    • one year ago
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    So we'll be back on track!?

  102. pooja195
    • one year ago
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    yes

  103. mathmate
    • one year ago
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    11.4 Multiplying and dividing rational expresssions =================================================== calculate \(\Large \frac{6}{15}\times \frac{12}{9}\)

  104. mathmate
    • one year ago
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    I would like you to treat the numbers as products of factors, and see if you can simplify without the calculator.

  105. pooja195
    • one year ago
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    i cant do it without a calc! :O

  106. mathmate
    • one year ago
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    It's not the answer we want, just the process. I'll do that for you.

  107. mathmate
    • one year ago
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    \Large \frac{2.3}{3.5}\times \frac{2.2.3}{3.3}

  108. mathmate
    • one year ago
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    \(\Large \frac{6}{15}\times \frac{12}{9}\) =\(\Large \frac{2.3}{3.5}\times \frac{2.2.3}{3.3}\) now cancel factors =\(\Large \frac{2}{5}\times \frac{2.2}{3}\) =\(\Large \frac{8}{15}\)

  109. mathmate
    • one year ago
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    If you replace each prime factor (2,3,5,...) by a polynomial, this is exactly what we do to multiply rational expressions!

  110. pooja195
    • one year ago
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    hmm ok

  111. mathmate
    • one year ago
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    simplify \(\Large \frac{x}{3x^2-9x}\times\frac{x-3}{2x^2+x-3}\) [do not forget the conditions whenever you cancel]

  112. pooja195
    • one year ago
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    \[\frac{ (x-3) }{ 3(x−1)(2x+3)(x−3) }\]

  113. pooja195
    • one year ago
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    correct so far?

  114. mathmate
    • one year ago
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    exactly, continue!

  115. pooja195
    • one year ago
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    \[\frac{ 1 }{ 3(x-1)(2x+3) }\] x/= 3

  116. mathmate
    • one year ago
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    Almost. The last step is all correct. There was a mistake in the first step:

  117. mathmate
    • one year ago
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    Oh, actually, you were right, I was wrong. The final answer is good! Congrats!

  118. pooja195
    • one year ago
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    O_o

  119. mathmate
    • one year ago
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    Do you still remember how to do a division of fractions?

  120. mathmate
    • one year ago
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    I'd say: flip the fraction after the \(\div\) sign.

  121. mathmate
    • one year ago
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    Do you agree?

  122. pooja195
    • one year ago
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    Can we keep doing multiplication ?

  123. mathmate
    • one year ago
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    Sure! Another one coming up

  124. mathmate
    • one year ago
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    \(\Large \frac{c^2-64}{4c^3} \times \frac{c}{c^2+9c+8}\)

  125. mathmate
    • one year ago
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    Simplify!

  126. pooja195
    • one year ago
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    \[\frac{ (c+8)(c−8) }{ 4cc(c+1)(c+8) }\]

  127. pooja195
    • one year ago
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    correct so faR?

  128. mathmate
    • one year ago
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    Excellent, go on!

  129. mathmate
    • one year ago
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    Remember you cancelled a "c" already. Keep track of that.

  130. pooja195
    • one year ago
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    Can you finish it off ;-;

  131. mathmate
    • one year ago
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    \(\Large \frac{c^2-64}{4c^3} \times \frac{c}{c^2+9c+8}\) =\(\Large \frac{(c+8)(c-8)}{4c^3} \times \frac{c}{(c+8)(c+1)}\) cancel c and (c+8) =\(\Large \frac{(c-8)}{4c^2(c+1)}~~~ x\ne -8,0\)

  132. mathmate
    • one year ago
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    Is the last part ok for you now?

  133. pooja195
    • one year ago
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    yes :)

  134. mathmate
    • one year ago
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    Another multiplication?

  135. pooja195
    • one year ago
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    nope ;-;

  136. mathmate
    • one year ago
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    Division: \(\Large \frac{1}{2}\div\frac{1}{5}\) =\(\Large \frac{1}{2}\times\frac{5}{1}\) Is this familiar to you?

  137. pooja195
    • one year ago
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    yes

  138. mathmate
    • one year ago
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    We would do the same with rational functions.

  139. mathmate
    • one year ago
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    Gimme a second to type it up.

  140. mathmate
    • one year ago
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    simplify \(\Large \frac{x}{x+6} \div \frac{x+3}{x^2-36}\)

  141. pooja195
    • one year ago
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    \[\frac{ x }{ x+6 }\times \frac{ (x+6)(x-6) }{( x+3) }\] \[\frac{ x(x-6) }{ (x+3) }~~~~~x \neq6\]

  142. mathmate
    • one year ago
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    Perfect!!!! Can't be better!

  143. pooja195
    • one year ago
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    REALLY? :O

  144. pooja195
    • one year ago
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    I did that without a calc :O

  145. mathmate
    • one year ago
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    Oops, what about the previous ones ! LOL

  146. pooja195
    • one year ago
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    smh

  147. mathmate
    • one year ago
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    Ready for another?

  148. pooja195
    • one year ago
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    yes!: )

  149. mathmate
    • one year ago
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    Simplify \(\Large \frac{45x^3-9x^2}{x} \div \frac{6(x-5)}{2}\)

  150. pooja195
    • one year ago
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    \[\frac{ 3x(5x−1) }{ x-5 }\]

  151. mathmate
    • one year ago
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    Wow, what did you eat for supper?

  152. pooja195
    • one year ago
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    xD

  153. mathmate
    • one year ago
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    Another?

  154. pooja195
    • one year ago
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    No i have this : ) next case!

  155. mathmate
    • one year ago
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    k. coming right up!

  156. mathmate
    • one year ago
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    11.5 Adding and Subtracting with like denominators =================================

  157. mathmate
    • one year ago
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    Like denominators mean that the denominators are either the same, or just a simple multiple.

  158. mathmate
    • one year ago
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    So it's a matter of adding the numerators, like:|dw:1433721592963:dw|

  159. pooja195
    • one year ago
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    ok

  160. mathmate
    • one year ago
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    now gimme a minute.

  161. mathmate
    • one year ago
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    Add \(\Large \frac{x+2}{x} + \frac{3x-2}{x}\)

  162. pooja195
    • one year ago
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    \[\frac{ 4x }{ x }\]

  163. mathmate
    • one year ago
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    ...and the next step is...

  164. pooja195
    • one year ago
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    the answer would be 4

  165. pooja195
    • one year ago
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    Because the x's cancel right?

  166. mathmate
    • one year ago
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    Yes, ...and...

  167. pooja195
    • one year ago
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    and??

  168. mathmate
    • one year ago
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    x\(\ne\)0 "conditions whenever cancelling a factor in the denominator".

  169. pooja195
    • one year ago
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    i knew it! >:(

  170. mathmate
    • one year ago
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    haha! Another?

  171. pooja195
    • one year ago
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    Next case!

  172. mathmate
    • one year ago
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    \(\Large \frac{3x-4}{x-4} - \frac{2x}{x-4}\)

  173. mathmate
    • one year ago
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    Just curious!

  174. pooja195
    • one year ago
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    i knew it was coming :P \[\frac{ 1x-4 }{ x-4}\]

  175. mathmate
    • one year ago
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    recall 1y = y

  176. pooja195
    • one year ago
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    1

  177. mathmate
    • one year ago
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    and..

  178. pooja195
    • one year ago
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    and??

  179. mathmate
    • one year ago
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    x\(\ne\)4

  180. mathmate
    • one year ago
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    The conditions are the only difficult part in these examples!

  181. pooja195
    • one year ago
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    ugh

  182. mathmate
    • one year ago
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    |dw:1433722130469:dw|

  183. pooja195
    • one year ago
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    multiplication right?

  184. mathmate
    • one year ago
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    Nope, perimeter would be the sum of all four sides.

  185. pooja195
    • one year ago
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    Next case!

  186. mathmate
    • one year ago
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    Please!

  187. mathmate
    • one year ago
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    We only have two more cases.

  188. pooja195
    • one year ago
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    I dont like this one ;-;

  189. pooja195
    • one year ago
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    im not even sure where to start .-.

  190. mathmate
    • one year ago
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    |dw:1433722487080:dw|

  191. mathmate
    • one year ago
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    can you finish it?

  192. pooja195
    • one year ago
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    \[\huge~\frac{ 2(x−2)(x^3+4x^2−12x+4) }{ (x+2)(x−2)(x−2) }\]

  193. mathmate
    • one year ago
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    wow, not like that. The addition had already been done, because the denominators are the same.

  194. mathmate
    • one year ago
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    All you need to do is to cancel (x-2) and give the answer!

  195. mathmate
    • one year ago
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    |dw:1433722913558:dw|

  196. pooja195
    • one year ago
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    >.<

  197. mathmate
    • one year ago
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    Do you see what was happening? I multiplied by 2 because the opposite sides are identical. I then add them, using the common denominator that we are given with. Did the additions to come to 2(x+4)(x-2)/(x-2) Since (x-2) is a common factor, we can cancel as long as x-2\(\ne\)0, or x\(\ne\)2 So the answer is 2(x+4), with x\(\ne\)2

  198. pooja195
    • one year ago
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    Got it

  199. mathmate
    • one year ago
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    good!

  200. mathmate
    • one year ago
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    another?

  201. pooja195
    • one year ago
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    no ;;

  202. mathmate
    • one year ago
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    ok, next case: 11.6 adding and subtracting rational functions ith unlike denominators ============================================= It's almost the same as 11.5. It's just you need to do dancing class with the denominator before adding and subtracting.

  203. mathmate
    • one year ago
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    example Find common denominator of \(\Large \frac{1}{12x} +\frac{2+x}{40x^4}\) The common denominator can be found similar to the dancing class: |dw:1433723859840:dw|

  204. mathmate
    • one year ago
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    Once you can find the common denominator readily, rest is relatively straight-forward.

  205. pooja195
    • one year ago
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    Makes sense :)

  206. mathmate
    • one year ago
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    So that's good, we'll proceed with an example.

  207. mathmate
    • one year ago
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    Simplify \(\Large \frac{x+1}{5} + \frac{2x}{6}\) Can you first try to find the common denominator?

  208. mathmate
    • one year ago
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    |dw:1433726621112:dw|