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pooja195

  • one year ago

@mathmate

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  1. mathmate
    • one year ago
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    Yep. Do you know direct and inverse variation?

  2. pooja195
    • one year ago
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    no ._.

  3. mathmate
    • one year ago
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    Direct variation is a function that varies directly with x. For example X Y 1 2 2 4 3 6 4 8 5 10 is a direct variation. Notice that Y/X is always the same ratio (of 2). so far so good?

  4. pooja195
    • one year ago
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    oh ok :)

  5. mathmate
    • one year ago
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    On a graph, direct variation is a straight line passing through the origin. |dw:1432866053517:dw| Here f1(x), f2(x), f3(x) are all direct variations, but g(x) is not, because it does not pass through the origin.

  6. mathmate
    • one year ago
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    ok with direct?

  7. pooja195
    • one year ago
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    yes :)

  8. mathmate
    • one year ago
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    For inverse variations, they work differently. Y is the inverse (or reciprocal) of x. This way, X*Y is a constant (same number). Example: X Y 1 2 2 1 3 2/3 4 1/2 5 2/5 6 1/3 ... notice the product of X*Y is always 2. the graph looks like this |dw:1432866347703:dw|

  9. mathmate
    • one year ago
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    ok for inverse variation?

  10. pooja195
    • one year ago
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    Yes i get that :)

  11. mathmate
    • one year ago
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    Can we move on the rational expressions?

  12. mathmate
    • one year ago
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    * to

  13. pooja195
    • one year ago
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    yes

  14. mathmate
    • one year ago
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    A rational expression is a polynomial divided by another polynomial.

  15. mathmate
    • one year ago
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    Example, (5x+3)/(2x+1) is a rational expression.

  16. mathmate
    • one year ago
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    The sum of two rational expressions is also a rational expression, just like the sum of two fractions is still a fraction.

  17. mathmate
    • one year ago
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    Even with one single rational expression, we can simplify , for example, what would be \(\large \frac{(4x+3)(x-1)}{(x-1)}\) ?

  18. pooja195
    • one year ago
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    i just foil right?

  19. mathmate
    • one year ago
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    Before you do foil, did you notice anything?

  20. pooja195
    • one year ago
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    we can cancel

  21. mathmate
    • one year ago
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    Yes, only as long as (x-1) does not equal zero!

  22. pooja195
    • one year ago
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    Yes :)

  23. mathmate
    • one year ago
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    So we say: \(\large \frac{(4x+3)(x-1)}{(x-1)}=4x+3\) if x\(\ne\)1

  24. mathmate
    • one year ago
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    ok with that?

  25. pooja195
    • one year ago
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    Yes :D

  26. mathmate
    • one year ago
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    So far, we've been lucky. Both the numerator and denominator contained (x-1) as a common factor. What if we are given the following to simplify: \(\large \frac{4x^2-x-3}{x-1}\) ?

  27. mathmate
    • one year ago
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    We cannot tell off-hand if there is any common factor, so what do we do?

  28. pooja195
    • one year ago
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    We use the wheel thingy

  29. mathmate
    • one year ago
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    Good, or "factorize" is the other word for the wheel thingy! Can you do that, please?

  30. pooja195
    • one year ago
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    |dw:1432867194095:dw| \[(4x^2-4x)+(3x-3)\] \[4x(x-1)+3(x-1)\] \[(4x+3)(x-1)\]

  31. mathmate
    • one year ago
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    Excellent! Now you replace the numerator of the rational expression with what you've got, what do you get? \(\large \frac{4x^2-x-3}{x-1}= ?\)

  32. pooja195
    • one year ago
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    \[\frac{ (4x+3)(x-1) }{ (x-1) }\] (x-1) cancel out. Final answer: \[ \huge 4x+3\]

  33. mathmate
    • one year ago
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    Whenever you cancel out in a rational expression, you need to ADD a condition that what you cancel out is not zero, so you need to add x\(\ne\)1 (or x-1\(\ne\)0).

  34. mathmate
    • one year ago
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    We don't do that in fractions because we almost always work with non-zero factors. If we had a fraction 5*0/4*0, cancelling the common factor 0 would give 5/4 as the answer, which is not the same as 0/0.

  35. mathmate
    • one year ago
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    So the answer is again 4x+3 with x\(\ne\)1.

  36. pooja195
    • one year ago
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    I dont like that :/

  37. mathmate
    • one year ago
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    the condition?

  38. pooja195
    • one year ago
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    Yep

  39. mathmate
    • one year ago
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    Yes, it complicates life a little, but that's the only difference from fractions.

  40. mathmate
    • one year ago
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    So whenever you need to simplify a rational expression, factorize as much as you can, and cancel as much as you can. Add one condition whenever you cancel a factor (unless the same factor appears twice) Example: (x+3)(x+1)^2/((x+1)^2(x+5)) =(x+3)/(x+5) if x\(\ne\)-1

  41. mathmate
    • one year ago
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    Are we good so far (except you don't like the condition! :) )

  42. pooja195
    • one year ago
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    yes :)

  43. mathmate
    • one year ago
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    If we multiply rational functions, it's the same as in fractions. Again, if we cancel, we need to specify conditions! \(\large \frac{x+5}{x-2} * \frac{x-2}{x+3} = \frac{x+5}{x+3} \) if x\(\ne\)2

  44. mathmate
    • one year ago
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    ok so far?

  45. pooja195
    • one year ago
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    yes.

  46. mathmate
    • one year ago
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    For divisions, it's the same as in multiplication, except that we need to flip the divisor and multiply. Example: \(\large \frac{x+5}{x-2} \div \frac{x+5}{x+3} = \frac{x+5}{x-2} * \frac{x+3}{x+5} =\frac{x+3}{x-2}\) if x\(\ne\)-5

  47. mathmate
    • one year ago
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    ok so far?

  48. pooja195
    • one year ago
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    yes :)

  49. mathmate
    • one year ago
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    Now for adding and subtracting with like denominators, it would be just like in fractions. Should I give an example anyway?

  50. pooja195
    • one year ago
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    Can we just skip those :P i know those

  51. mathmate
    • one year ago
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    What about with unlike denominators, are you ok with those?

  52. pooja195
    • one year ago
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    no...

  53. mathmate
    • one year ago
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    ok. But back to those with like denominators: after adding or subtracting, you will still need to factorize and look for factors to cancel AND add a note/condition. Is that ok?

  54. pooja195
    • one year ago
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    yes

  55. mathmate
    • one year ago
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    Ok, now +/- expressions with unlike denominators. Just like 1/3+2/5 common denominator is 3*5=15 so we'll make equivalent fractions before adding = 5/15 + 6/15 now add =11/15 ok?

  56. pooja195
    • one year ago
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    yes

  57. mathmate
    • one year ago
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    Now for rational exprssions: \(\large \frac{x+1}{x-2}-\frac{x-2}{x+4}\)

  58. mathmate
    • one year ago
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    What would be the common denominator? (do not FOIL, otherwise we need to factorize afterwards.)

  59. pooja195
    • one year ago
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    x+4

  60. mathmate
    • one year ago
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    We need the product of the two denominators, (x-2)(x+4), so that we can make equivalent expressions. \(\large \frac{x+1}{x-2}-\frac{x-2}{x+4}\) =\(\large \frac{(x+1)(x+4)}{(x-2)(x+4)}-\frac{(x-2)(x-2)}{(x+4)(x-2)}\)

  61. mathmate
    • one year ago
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    ok?

  62. mathmate
    • one year ago
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    To this we have to add the conditions that x-2\(\ne\)0 and x+4\(\ne\)0

  63. mathmate
    • one year ago
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    ok so far in making equivalent fractions?

  64. mathmate
    • one year ago
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    The rest of it is adding two rational expressions with like denominators. After addition/subtraction, you need to factorize and cancel expression (and add conditions if applicable).

  65. pooja195
    • one year ago
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    im so confused and lost T_T

  66. mathmate
    • one year ago
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    @pooja195 xD

  67. pooja195
    • one year ago
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    NO!! T_T

  68. mathmate
    • one year ago
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    Yesterday we went through 11.2 to 11.6. Now it's time to do some practice!

  69. pooja195
    • one year ago
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    next class

  70. mathmate
    • one year ago
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    :( your way out!

  71. mathmate
    • one year ago
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    Just kidding, sure, later!

  72. mathmate
    • one year ago
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    @pooja195 Time to work!

  73. pooja195
    • one year ago
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    O_O

  74. pooja195
    • one year ago
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    Just remebered i have a thing to do.... oops cant work :P

  75. mathmate
    • one year ago
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    Ok, I'll grab you later, ok?

  76. pooja195
    • one year ago
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    BWAHAHAHA XDDD :3 yr so innocent xS

  77. pooja195
    • one year ago
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    im free :# we can math :P

  78. mathmate
    • one year ago
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    ok! I was O_o

  79. mathmate
    • one year ago
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    R U good with 11.3?

  80. pooja195
    • one year ago
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    :/

  81. mathmate
    • one year ago
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    Want a practice to make sure?

  82. pooja195
    • one year ago
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    yes

  83. mathmate
    • one year ago
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    BTW, were the correct answers in the quiz all between 11.2 and 11.3?

  84. pooja195
    • one year ago
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    no :/

  85. mathmate
    • one year ago
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    So you did get some in other sections, that's good!

  86. pooja195
    • one year ago
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    ummm no

  87. mathmate
    • one year ago
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    Now try to simplify: \(\large f(x)=\frac{x^2+x-6}{x^2-x-12}\) Do not forget to mention the condition, if any

  88. pooja195
    • one year ago
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    i thought i did great on it but i ended up failing

  89. pooja195
    • one year ago
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    ;-; can we start with something easier :(

  90. mathmate
    • one year ago
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    Ok, simplify \(\large f(x)=\frac{x^2+x-6}{(x+3)(x-4)}\) :)

  91. pooja195
    • one year ago
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    |dw:1432933988455:dw| IDK T_T

  92. mathmate
    • one year ago
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    Remember, when the product (-6) is negative, the two numbers you're looking for have different signs. Does that help?

  93. mathmate
    • one year ago
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    |dw:1432934285205:dw| Which pair works?

  94. pooja195
    • one year ago
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    3 -2 .-.

  95. pooja195
    • one year ago
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    (x+3)(x-2)

  96. mathmate
    • one year ago
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    Exactly. Are you able to complete the answer now?

  97. pooja195
    • one year ago
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    \[\frac{ (x-2)}{ (x+4) }\]

  98. mathmate
    • one year ago
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    Almost, just a transcription error (should be x-4 at the bottom) AND missing very important information... Do you know what's missing?

  99. pooja195
    • one year ago
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    no

  100. mathmate
    • one year ago
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    The condition that (x+3)\(\ne\)0. Teachers are waiting to jump on that! :(

  101. mathmate
    • one year ago
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    So the complete answer is (x-2)/(x-4) with x\(\ne\) -3

  102. mathmate
    • one year ago
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    ok so far?

  103. pooja195
    • one year ago
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    I hate that

  104. pooja195
    • one year ago
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    ye the rest is fine :)

  105. mathmate
    • one year ago
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    Just have to remember that whenever you cancel a term, it comes with a condition. Like when you buy candies, you always pay! :)

  106. mathmate
    • one year ago
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    Want another one similar?

  107. pooja195
    • one year ago
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    no :(

  108. mathmate
    • one year ago
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    T_T !

  109. pooja195
    • one year ago
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    ok fine another one

  110. mathmate
    • one year ago
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    ok, here's another one. Notice that the numerator is the same as the previous, so you are allowed to reuse your own work this time. Simplify \(\large \frac{x^2+x-6}{x^2-2x-15}\)

  111. pooja195
    • one year ago
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    |dw:1432935523037:dw| \[\frac{(x-2)(x+3) }{ (x-5)(5+3)}\] Final answer \[\frac{ (x-2) }{ (x+5) }\]

  112. mathmate
    • one year ago
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    oh...oh

  113. pooja195
    • one year ago
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    O_O ?

  114. mathmate
    • one year ago
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    Remember the last problem? transcription and something missing???

  115. pooja195
    • one year ago
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    T_T i dont know this!!! >:(

  116. mathmate
    • one year ago
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    Good up to this: \(\frac{(x-2)(x+3) }{ (x-5)(x+3)}\) After that, the final answer after cancelling is \(\frac{(x-2)}{ (x-5)}\) with the condition that x\(\ne\)5.

  117. mathmate
    • one year ago
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    @pooja195

  118. pooja195
    • one year ago
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    :)

  119. mathmate
    • one year ago
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    We were on 11.3

  120. mathmate
    • one year ago
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    All you have to remember is if you get candies, you pay!

  121. mathmate
    • one year ago
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    candies ~ cancel

  122. pooja195
    • one year ago
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    :( i hate that

  123. mathmate
    • one year ago
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    Sorry, we have to stick to rules! Now can you simplify (x-2)(x+4)/(x+4) ?

  124. pooja195
    • one year ago
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    (x-2)

  125. mathmate
    • one year ago
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    You cancelled (x+4) right?

  126. pooja195
    • one year ago
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    yea....

  127. mathmate
    • one year ago
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    So we write (x-2).....

  128. pooja195
    • one year ago
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    thats what i did .-.

  129. mathmate
    • one year ago
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    You were close: (x-2) for x\(\ne\)2 is the complete answer.

  130. mathmate
    • one year ago
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    * for x\(\ne\) -4

  131. pooja195
    • one year ago
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    >:(

  132. mathmate
    • one year ago
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    ok, let's try another: simplify (x-4)(x+2)/(x-4)

  133. pooja195
    • one year ago
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    (x+2)

  134. mathmate
    • one year ago
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    Almost, it would be (x+2) for x\(\ne\)4 because if x=4, the expression/function is 0/0, and that does not have a value. So our answer of (x+2) is valid as long as x\(\ne\) 4

  135. pooja195
    • one year ago
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    Ahhh >:( T_T

  136. mathmate
    • one year ago
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    Now try this: simplify (2x+5/2)(x+3)/(2x+5/2)

  137. pooja195
    • one year ago
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    No no more of these T_T

  138. mathmate
    • one year ago
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    I am looking forward to see you do one completely by yourself, because I won't be there to help you when you do your quiz!

  139. pooja195
    • one year ago
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    x+3 ;-;

  140. mathmate
    • one year ago
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    The answer is (x+3) for 2x+5/2\(\ne\)0 Remember when 2x+5/2=0, the expression becomes 0/0 and is undefined. So to avoid this situation, we specify that 2x+5/2\(\ne\)0

  141. pooja195
    • one year ago
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    Can we please skip these T_T

  142. mathmate
    • one year ago
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    Sorry, math is a cumulative knowledge. If we don't do it right, all the following topics will be wrong! I am asking you to show me at least once that you will follow a cancellation with a condition.

  143. mathmate
    • one year ago
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    Simplify (2x+3)(x-1)/(x-1)

  144. pooja195
    • one year ago
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    2x+3 x/= 1??????

  145. mathmate
    • one year ago
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    yes, put if, for, for all, or whenever... to indicate that it is a condition. so 2x+3 for all x\(\ne\) 1

  146. mathmate
    • one year ago
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    Are you ok now?

  147. mathmate
    • one year ago
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    I will skip this step and move on to 11.4. But remember that in all the other topics, whenever you cancel, you still have to write the condition, else the answer is wrong without it. Are we good?

  148. pooja195
    • one year ago
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    Yes

  149. mathmate
    • one year ago
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    Ready for 11.4?

  150. pooja195
    • one year ago
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    Yes

  151. mathmate
    • one year ago
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    @pooja195

  152. pooja195
    • one year ago
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    I iz here

  153. mathmate
    • one year ago
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    11.4 is on rational expressions and not radical expression, I suppose. Can you confirm?

  154. pooja195
    • one year ago
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    yea lol rational is right

  155. mathmate
    • one year ago
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    ok, try this: simplify \(\large \frac{x-2}{x+3}\times\frac{x+3}{x-4}\)

  156. pooja195
    • one year ago
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    (x-2)(x+3)/(x+3)(x-4) \[\huge \frac{ x-2 }{ x-4 }\neq 3\]

  157. mathmate
    • one year ago
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    That's the idea! We write the condition separate though, \(\large \frac{ x-2 }{ x-4 }\ for\ all\ x\neq 3\)

  158. mathmate
    • one year ago
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    correction: \(\large \frac{ x-2 }{ x-4 }\ for\ all\ x\neq -3\) because we don't want x+3=0

  159. mathmate
    • one year ago
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    Now, try Simplify \(\large \frac{x-2}{5x+6}\times\frac{2x-3}{x-2}\)

  160. pooja195
    • one year ago
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    (x-2)(x-2)/(5x+6)(2x-3) T_T Lets do a diffrent one! :|

  161. mathmate
    • one year ago
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    You need to cancel the (x-2) at the top with the (x-2) at the bottom, the will give (2x-3)/(5x+6) for all x\(\ne\) 2

  162. mathmate
    • one year ago
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    ok try this: Simplify \(\large \frac{x^2-5x+6}{5x+6}\times\frac{5x+3}{x-2}\)

  163. pooja195
    • one year ago
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    (x-3)(x-2)(5x+3)/(5x+6)(x-2) \[\frac{ (x-3)(5x+3) }{ (5x+6)(x-2) }\neq-2\]

  164. mathmate
    • one year ago
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    you cancelled the x-2 at the bottom, so x-2\(\ne\)0, or x\(\ne\)2 The answer will then read: \(\frac{ (x-3)(5x+3) }{ (5x+6)} \ for\ all\ x\neq2\)

  165. pooja195
    • one year ago
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    >_<

  166. mathmate
    • one year ago
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    Simplify \(\large \frac{x-3}{5x+6}\div \frac{x-2}{5x-6}\)

  167. mathmate
    • one year ago
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    You'll need to invert the second term to change division to multiplication: =\(\large \frac{x-3}{5x+6}\times \frac{5x-6}{x-2}\)

  168. mathmate
    • one year ago
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    oops the question is wrong. It is meant to have 5x+6 top and bottom! \(\large \frac{x-3}{5x+6}\div \frac{x-2}{5x+6}\) =\(\large \frac{x-3}{5x+6}\times \frac{5x+6}{x-2}\)

  169. pooja195
    • one year ago
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    O_o

  170. mathmate
    • one year ago
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    Now we cancel the common factor 5x+6 (as long as 5x+6\(\ne\)0. so we write =\(\large \frac{x-3}{x-2}\) for all 5x+6\(\ne\)0 =\(\large \frac{x-3}{x-2}\) for all x\(\ne\)-6/5

  171. pooja195
    • one year ago
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    ok i kinda get it

  172. mathmate
    • one year ago
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    @pooja195 we were at 11.4 are you ready?

  173. pooja195
    • one year ago
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    ye

  174. mathmate
    • one year ago
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    It seems better now. Are you good with multiplication and division of rationals?

  175. pooja195
    • one year ago
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    kind of :/

  176. mathmate
    • one year ago
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    ok, we'll go through all sections, and then come back to those which you feel wobbly. BTW, do you have some examples from school, we can work on those.

  177. pooja195
    • one year ago
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    yeah lol my hoe work ill put up a problem from that : )

  178. mathmate
    • one year ago
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    Sure! It will be less boring.

  179. pooja195
    • one year ago
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    xD \[\frac{ x^2+2x-15 }{ 3x^2+6x }\div \frac{ 3x^2-9x }{ x^2+7x+10 }\]

  180. mathmate
    • one year ago
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    So what would be your first step?

  181. pooja195
    • one year ago
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    factor?

  182. mathmate
    • one year ago
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    That's ok, but I would prefer to put the second term upside down, so we will always working with multiplication.

  183. mathmate
    • one year ago
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    * always be...

  184. pooja195
    • one year ago
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    ooo right flip

  185. mathmate
    • one year ago
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    \(\frac{ x^2+2x-15 }{ 3x^2+6x }\div \frac{ x^2+7x+10 }{ 3x^2-9x }\)

  186. mathmate
    • one year ago
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    Oops too small, but I can still read it.

  187. pooja195
    • one year ago
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    lol ok so now we factor right?

  188. mathmate
    • one year ago
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    right, lemme fix the div sign, but you can go ahead and factor. Remember the first step is to take out the common factors of each expression, ex. in 3x^2+6x. Take out 3x.

  189. mathmate
    • one year ago
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    \(\large \frac{ x^2+2x-15 }{ 3x(x+2) }\times \frac{ x^2+7x+10 }{ 3x^2-9x }\)

  190. pooja195
    • one year ago
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    Ok this is the part where i get kinda confused....you know how theres like two seperate equations on top? They both can be factored but how would i write it out?

  191. mathmate
    • one year ago
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    Actually on second look, would the original question be a multiplication instead of a division?

  192. mathmate
    • one year ago
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    You can start with writing them separately in two separate numerators. After that, we cancel.

  193. mathmate
    • one year ago
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    But it seems that there is not much to cancel unless the original question was a multiplication.

  194. pooja195
    • one year ago
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    its division :/

  195. mathmate
    • one year ago
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    Sorry, I take it back. There is at least one that we can cancel.

  196. mathmate
    • one year ago
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    So you can go ahead and factor, as though they are two separate terms. We can combine them after cancelling.

  197. pooja195
    • one year ago
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    huh :?

  198. mathmate
    • one year ago
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    I mean you can proceed with this: \(\large \frac{ x^2+2x-15 }{ 3x(x+2) }\times \frac{ x^2+7x+10 }{ 3x^2-9x }\)

  199. pooja195
    • one year ago
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    How would i write out the factors?

  200. mathmate
    • one year ago
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    The bottom one on the right should take out 3x as well, so you put \(\large \frac{ x^2+2x-15 }{ 3x(x+2) }\times \frac{ x^2+7x+10 }{ 3x(x-3) }\)

  201. mathmate
    • one year ago
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    For the ones in the numerator, you can put them where they are now.

  202. mathmate
    • one year ago
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    If you want, we can factor the left numerator together.

  203. mathmate
    • one year ago
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    \(\large x^2+2x-15\) =\(\large (x- \ )(x+ \ )\)

  204. pooja195
    • one year ago
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    ok |dw:1433012589434:dw| |dw:1433012632806:dw| \[\frac{ (x+5)(x-3) }{ 3x(x+2) } \times \frac{ (x+5)(x+2) }{ 3x(x-3) }\]

  205. mathmate
    • one year ago
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    Good job! Speed of a bullet!

  206. mathmate
    • one year ago
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    can you now post me the factors that you can cancel, and write down the condition associated with each one?

  207. mathmate
    • one year ago
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    like: (x-3) means x\(\ne\)3

  208. mathmate
    • one year ago
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    You can write all the factors on top as one single numerator, and similarly for the denominators. Looks like this:

  209. pooja195
    • one year ago
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    \[\frac{ (x+5)^2 }{ 3x }\]

  210. mathmate
    • one year ago
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    Yes, but there are two "3x" , so you write 9x^2 at the bottom.

  211. mathmate
    • one year ago
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    ...and don't forget to add the 2 conditions that correspond to the two factors that you cancelled out.

  212. pooja195
    • one year ago
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    >:(

  213. pooja195
    • one year ago
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    \[x \neq-3~~~~~~~x \neq2\]

  214. mathmate
    • one year ago
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    yes. Please write the answer with the conditions on one line, the conditions are "and" because both have to be satisfied.

  215. mathmate
    • one year ago
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    If you want to see how it should be done, see the example near the bottom of the page: http://www.purplemath.com/modules/rtnlmult.htm

  216. pooja195
    • one year ago
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    Does it have to be written like that? :/

  217. mathmate
    • one year ago
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    It would be the simplest, simpler than what I would have done.

  218. pooja195
    • one year ago
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    \[\frac{ (x+5)^2 }{ 3x },x \neq2,x \neq-3\]

  219. mathmate
    • one year ago
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    That's ok too, but do remember it's 9x^2 at the bottom, because we had two 3x left at the bottom!

  220. pooja195
    • one year ago
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    oo right >_< oops

  221. mathmate
    • one year ago
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    \(\large\frac{ (x+5)^2 }{ 9x^2 },x \neq2,x \neq-3\)

  222. mathmate
    • one year ago
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    Excellent, you got one done, and it wasn't a simple one like my (boring) ones.

  223. pooja195
    • one year ago
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    XD

  224. mathmate
    • one year ago
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    Got other ones to try?

  225. pooja195
    • one year ago
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    \[\huge \frac{ x^2-64 }{ 3x^3 }by~(8-8)\]

  226. mathmate
    • one year ago
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    what's "by" O_o

  227. mathmate
    • one year ago
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    Sure it's (8-8) ? (equals zero!) O_o!

  228. pooja195
    • one year ago
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    \[\huge \frac{ (x+8)(x-8) }{ 3x^3}\div \frac{ 1 }{ (x-8) }\]

  229. mathmate
    • one year ago
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    Ok! I think you did the division twice (you flipped, and you kept the division sign!)

  230. mathmate
    • one year ago
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    Is it \(\huge \frac{ (x+8)(x-8) }{ 3x^3}\times \frac{ 1 }{ (x-8) }\)

  231. mathmate
    • one year ago
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    And you factored the top right-away, ;) ?

  232. pooja195
    • one year ago
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    yes :) xD

  233. mathmate
    • one year ago
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    Excellent!

  234. mathmate
    • one year ago
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    What's your answer then?

  235. mathmate
    • one year ago
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    Bet you this one comes out in the quiz.

  236. pooja195
    • one year ago
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    \[\huge\frac{ 1(x+8) }{ 3x^3 },x \neq -8\]

  237. mathmate
    • one year ago
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    Another one done, with 100%.

  238. pooja195
    • one year ago
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    :) MORE!!!:D

  239. mathmate
    • one year ago
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    yes please! More, more....

  240. pooja195
    • one year ago
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    You pick :P ;)

  241. mathmate
    • one year ago
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    I like your examples better, they are less boring.

  242. mathmate
    • one year ago
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    ok, I'll pick one.

  243. pooja195
    • one year ago
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    LOL xD

  244. mathmate
    • one year ago
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    Simplify \(\large \frac{x^2+6x+9}{x^2-9}.\frac{3x-9}{x^2+2x-3}\)

  245. mathmate
    • one year ago
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    There is no grouping to be done in the factoring because the coefficients of the quadratic expressions are all one (1).

  246. pooja195
    • one year ago
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    \[\frac{ (x+3)(x+3) }{ (x+3)(x-3) }\times \frac{ 3x-9 }{x+3)(x-1) }\]

  247. mathmate
    • one year ago
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    Wow!

  248. mathmate
    • one year ago
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    3x-9 = 3(x-3)

  249. mathmate
    • one year ago
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    Did you use the perfect square and diff. of two squares?

  250. mathmate
    • one year ago
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    or you factored all three?

  251. pooja195
    • one year ago
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    Diffrence of square \[\huge \frac{ (x+3)(x+3) }{ (x+3)(x-3) } \times \frac{ 3(x-3) }{ (x+3)(x-1) }\] not sure how to do the next part

  252. mathmate
    • one year ago
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    Excellent. Now you cancel, and note the conditions as you cancel.

  253. mathmate
    • one year ago
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    (x-3) and (x+3)^2 will be cancelled top and bottom, leaving \(\large \frac{3}{x-1}\), x\(\ne\)-3,1

  254. mathmate
    • one year ago
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    * \(\large \frac{3}{x-1}\), x\(\ne\)-3,3 You don't have to worry about x+3 twice.

  255. mathmate
    • one year ago
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    ready for another?

  256. pooja195
    • one year ago
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    yes! :)

  257. mathmate
    • one year ago
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    Simplify \(\large \frac{2x^2+x-6}{x^2-2x-8} \div \frac{2x^2-x-3}{x^2-3x-4}\)

  258. pooja195
    • one year ago
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    \[\frac{ (2x-3)(x+2) }{ (x-4)(x+2) }\times \frac{ (2x-3)(x+1) }{ (x+4)(x-1) }\]

  259. mathmate
    • one year ago
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    Great, just you need to flip, and I think bottom left should read (x-4)(x+1) before flipping.

  260. pooja195
    • one year ago
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    :(

  261. mathmate
    • one year ago
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    That was a great piece of work, involving 4 factorizations,, out of which you got 3 of them perfect, and one was just a switch of the sign! :)

  262. mathmate
    • one year ago
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    * bottom-right

  263. pooja195
    • one year ago
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    wait so what should it look like :. ?

  264. mathmate
    • one year ago
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    The answer should be very simple, almost the simplest possible, plus three conditions.

  265. mathmate
    • one year ago
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    \(\large \frac{ (2x-3)(x+2) }{ (x-4)(x+2) }\times \frac{ (x-4)(x+1) }{ (2x-3)(x+1) }\) Can you finish it?

  266. mathmate
    • one year ago
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    *four conditions

  267. pooja195
    • one year ago
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    all of them cancel out! T_T

  268. mathmate
    • one year ago
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    That's good, so what's the answer? (it's not zero).

  269. pooja195
    • one year ago
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    \[\huge x~\neq-4~~~x~\neq2~~~~~x~\neq1~~~~~????\]

  270. mathmate
    • one year ago
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    Almost, compare 5/5=1 x-4\(\ne\)0 means x\(\ne\)4, etc. so \(\large 1, x\ne -2,-1,3/2,4\)

  271. mathmate
    • one year ago
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    Continue with these "easy" ones, or some addition?

  272. pooja195
    • one year ago
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    T_T

  273. pooja195
    • one year ago
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    i wanna move onto something more challenging :)

  274. mathmate
    • one year ago
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    Like addition and subtration?

  275. mathmate
    • one year ago
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    *subtraction

  276. pooja195
    • one year ago
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    ye

  277. mathmate
    • one year ago
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    ok, Simplify \(\large \frac{4}{x^2-16}+\frac{3}{x^2+8x+16}\)

  278. mathmate
    • one year ago
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    Step 1: factorize, then we can find the lowest common multiple (LCM)

  279. mathmate
    • one year ago
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    or the common denominator.

  280. pooja195
    • one year ago
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    \[\huge \frac{ 4 }{ (x+4)(x-4) }+\frac{ 3 }{(x+4)(x+4) }\]

  281. pooja195
    • one year ago
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    :/

  282. mathmate
    • one year ago
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    Excellent!

  283. mathmate
    • one year ago
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    now we have on denominator (x-4)(x+4) and the other (x+4)(x+4). Can we find the common denominator? Example: common denominator of 6 and 9: 6=2*3, 9=3*3, so common den. = 2*3*3 the common denominator contains all the factor of each of the originals.

  284. mathmate
    • one year ago
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    *one

  285. pooja195
    • one year ago
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    x+4

  286. mathmate
    • one year ago
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    x+4 does not contain x-4, and x+4 twice. Hint: the common denominator here has three factors.

  287. pooja195
    • one year ago
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    ;-; idk

  288. mathmate
    • one year ago
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    (x-1)(x+4)(x+4) contains (x-1)(x+4) which is the first one, and also contains (x+4)(x+4) the second one. So (x-1)(x+4)(x+4) is the common denominator. You can also get it by multiplying together the two denominators, but cancelling one of the repetitions. \((x-4)\color{red}{(x+4)} (x+4)\color{red}{(x+4)}\) We will keep only one of the two reds because they repeat. That gives \((x-4)\color{red}{(x+4)} (x+4)\)

  289. mathmate
    • one year ago
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    Now step 2: For the first term, we need to multiply top and bottom by (x+2) to get the bottom to be the common denominator, i.e. \(\large \frac{4\color{blue}{(x+4)}}{(x-4)(x+4)\color{blue}{(x+4)}}\)

  290. mathmate
    • one year ago
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    Similarly for the second term: \(\large \frac{3\color{blue}{(x-4)}}{\color{blue}{(x-4)}(x+4)(x+4)}\)

  291. mathmate
    • one year ago
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    Since we have a common factor, we can now add: \(\large \frac{4(x+4)}{(x-4)(x+4)(x+4)}+\frac{3(x-4)}{(x-4)(x+4)(x+4)}\)

  292. mathmate
    • one year ago
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    = \(\large \frac{4x+16+3x-12}{(x-4)(x+4)(x+4)}\) = \(\large \frac{7x+4}{(x-4)(x+4)(x+4)}\)

  293. mathmate
    • one year ago
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    and that's the answer.

  294. mathmate
    • one year ago
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    Did you follow all the steps?

  295. mathmate
    • one year ago
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    If you get stuck, tell me where.

  296. mathmate
    • one year ago
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    You can also read example 4 of the following link: http://www.cliffsnotes.com/math/algebra/algebra-ii/rational-expressions/adding-and-subtracting-rational-expressions

  297. mathmate
    • one year ago
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    For the common denominator part, I will rephrase as follows: You can also get it by multiplying together the two denominators, but cancelling one of the repetitions, if any, \(between\) the denominators. \([(x−4)\color{red}{(x+4)}] [(x+4)\color{red}{(x+4)}]\) We will keep only one of the two reds because they repeat. That gives \((x−4)(x+4)\color{red}{(x+4)}\) as the common denominator.

  298. pooja195
    • one year ago
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    oh ok i get it :)

  299. mathmate
    • one year ago
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    The next step is to make equivalent fractions by multiplying by appropriate factors to make the denominator the common denominator. This is done for each term to be added/subtracted.

  300. pooja195
    • one year ago
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    ok =)

  301. mathmate
    • one year ago
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    Try another one?

  302. pooja195
    • one year ago
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    No T_T

  303. mathmate
    • one year ago
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    Do you have anything from your school notes?

  304. mathmate
    • one year ago
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    Example or exercises?

  305. pooja195
    • one year ago
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    no o.o

  306. mathmate
    • one year ago
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    ok, shall we take a break/

  307. pooja195
    • one year ago
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    mmmmmmmmmmmmmmmm k :)

  308. pooja195
    • one year ago
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    No lets continue

  309. mathmate
    • one year ago
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    Do you have some examples for the easy ones (like denominators)?

  310. pooja195
    • one year ago
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    No. I'd prefer you give the problems... :/

  311. mathmate
    • one year ago
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    Ok, the boring ones!

  312. mathmate
    • one year ago
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    Simplify \(\large \frac{4x+3}{x^2-16}+\frac{3x+4}{x^2-16}\)

  313. mathmate
    • one year ago
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    The terms already have a common denominator, so we just add the numerators: \(\large \frac{4x+3\ \ +\ \ 3x+4}{x^2-16}\) Can you finish it?

  314. pooja195
    • one year ago
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    \[\frac{7\left(x+1\right)}{\left(x+4\right)\left(x-4\right)} \]

  315. mathmate
    • one year ago
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    Excellent!

  316. mathmate
    • one year ago
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    Simplify \(\large \frac{x^2-3x+2}{2x-4}-\frac{x-1}{2x-4}\)

  317. mathmate
    • one year ago
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    Like the one before, the two terms already have a common denominator, so just do the subtraction of the numerators and factor, if possible, the difference.

  318. pooja195
    • one year ago
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    \[\frac{\left(x-3\right)\left(x-1\right)}{2\left(x-2\right)} \]

  319. mathmate
    • one year ago
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    Oh yes, I thought the difference is not factorable! lol Great job!

  320. pooja195
    • one year ago
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    v_v next

  321. mathmate
    • one year ago
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    k

  322. mathmate
    • one year ago
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    \(\large \frac{2x+3}{x^2+4x+4} +\frac{x^2+2x+1}{(x+2)^2}\)

  323. pooja195
    • one year ago
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    ;-; not this one! Lets skip it! :D

  324. mathmate
    • one year ago
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    lol, don't get fooled! The two denminators are actually identical, if you factor the first or expand the second, you will find that they are the same. Just add the numerators and factor if necessary.

  325. pooja195
    • one year ago
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    \[\huge\frac{ 2x+3}{ (x+2)(x+2)}+\frac{ (x+1)(x+1) }{ (x+2)(x+2) }\]

  326. mathmate
    • one year ago
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    for me?

  327. pooja195
    • one year ago
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    \[\huge\frac{ 2x+3(x+1)^2}{ (x+2)(x+2) }\]

  328. mathmate
    • one year ago
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    Watch out: \(\huge\frac{ 2x+3\color{red}{+}(x+1)^2}{ (x+2)(x+2) }\)

  329. mathmate
    • one year ago
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    Keep going, you're almost there!

  330. mathmate
    • one year ago
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    You need to expand the (x+1)^2 using FOIL or identities and then add.

  331. pooja195
    • one year ago
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    \[\frac{ 2x+3+x^2+1x+1x+1 }{ (x+2)^2}=\frac{ x^2+4x+4 }{ (x+2) }=\frac{ (x+2)(x+2 }{ (x+2)(x+2)}\]

  332. mathmate
    • one year ago
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    Great! So after cancelling you have a numeric answer with a single condition:

  333. pooja195
    • one year ago
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    \[x \neq2\]

  334. mathmate
    • one year ago
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    The answer is 1 (5*5/(5*5)=1) and the condition is x+2\(\ne\)0, or x\(\ne\)-2

  335. mathmate
    • one year ago
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    Remember we don't want what we cancelled to be zero.

  336. pooja195
    • one year ago
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    got it.

  337. mathmate
    • one year ago
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    Try the unlike denominators now or later?

  338. mathmate
    • one year ago
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    I'll start with this: Add \(\large \frac{2}{9}+\frac{7}{12}\)

  339. mathmate
    • one year ago
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    Please show work!

  340. pooja195
    • one year ago
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    No lets skip this section i dont like this section

  341. mathmate
    • one year ago
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    ok, 9=3*3, 12= 3*4 Let's find the LCM: |dw:1433023297993:dw| so \(\large \frac{2}{9}+\frac{7}{12}=\frac{2*4}{9*4}+\frac{7*3}{12*3}=\frac{8}{36}+\frac{21}{36}=\frac{29}{36}\)

  342. mathmate
    • one year ago
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    @pooja195

  343. pooja195
    • one year ago
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    I IZ HERE

  344. mathmate
    • one year ago
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    hold on! have to find the question!

  345. pooja195
    • one year ago
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    ugh no i put everything away ;-; lets just use yours :(

  346. pooja195
    • one year ago
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    mathmate i thin your pm's are lagging

  347. pooja195
    • one year ago
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    *think

  348. pooja195
    • one year ago
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    omg

  349. mathmate
    • one year ago
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    forget the previous one with the wrong operator. Simplify \(\large \frac{2x^2+x-6}{x^2-2x-8}\div \frac{2x^2-x-3}{x^2-3x-4}\)

  350. pooja195
    • one year ago
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    omg

  351. mathmate
    • one year ago
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    First step: flip.

  352. mathmate
    • one year ago
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    \(\large \frac{2x^2+x-6}{x^2-2x-8}\times \frac{x^2-3x-4}{2x^2-x-3}\)

  353. mathmate
    • one year ago
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    Next, factor, one piece at a time, and show intermediate steps, please!

  354. pooja195
    • one year ago
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    \[\huge \frac{ 2x^2+x-6 }{ x^2-2x-8 }\times \frac{ x^2-3x-4 }{ 2x^2-x-3 }\]

  355. mathmate
    • one year ago
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    I would do it like that: For the top left, we have -12 and +1. We know 3*4=12 with a difference of 1 (diff because 12 is negative). So figure out the right signs, and that will be +4 -3 (to give +1).

  356. pooja195
    • one year ago
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    \[huge(x+4)(x-3)\] \[\huge(2x^2+4x)(-3x-6)\] \[\huge2x(x+2)~~~~~-3(x+2)\]

  357. mathmate
    • one year ago
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    Then proceed with grouping: 2x^2+4x -3x-6 = 2x(x+2) -3(x+2) = (2x-3)(x+2)

  358. mathmate
    • one year ago
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    Good job, continue please!

  359. pooja195
    • one year ago
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    [\huge \frac{ (2x-3)(x+2) }{ (x-4)(x+2) }\times \frac{(x-4)(x+1) }{ 2x^2-x-3 }\]

  360. pooja195
    • one year ago
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    \[(2x^2+2x)(-3x-3)\] \[2x(x+1)~~~~~~~-3(x+1)\] \[(2x-3)(x+1)\]

  361. mathmate
    • one year ago
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    You mean \[\huge \frac{ (2x-3)(x+2) }{ (x-4)(x+2) }\times \frac{(x-4)(x+1) }{ 2x^2-x-3 }\]

  362. pooja195
    • one year ago
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    \[[\huge \frac{ (2x-3)(x+2) }{ (x-4)(x+2) }\times \frac{(x-4)(x+1) }{ (2x-3)(x+1) }\] \]

  363. mathmate
    • one year ago
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    so far so good!

  364. mathmate
    • one year ago
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    Now the finishing touch!

  365. pooja195
    • one year ago
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    the answer is 1 .-.

  366. mathmate
    • one year ago
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    right, but not finished yet!

  367. pooja195
    • one year ago
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    omg no T_T the conditions

  368. mathmate
    • one year ago
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    The conditions can be obtained by equating each factor cancelled to zero. Ex. 2x-3\(\ne\)0 means x\(\ne\)3/2, etc.

  369. mathmate
    • one year ago
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    Didn't your teacher ask you to specify the conditions, or is it just I?

  370. pooja195
    • one year ago
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    no he doesnt make us list those :/

  371. mathmate
    • one year ago
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    Never?

  372. pooja195
    • one year ago
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    nope :/

  373. mathmate
    • one year ago
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    Even in the quiz? Is it multiple choice?

  374. mathmate
    • one year ago
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    Ok, do it just for this one,to remind you that they should be there. After that, if they are listed correctly, you don't have to do it until after the quiz!!

  375. pooja195
    • one year ago
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    ;-;

  376. mathmate
    • one year ago
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    So the four conditions are: 2x-3\(\ne\)0 x+2\(\ne\)0 x-4\(\ne\)0 x+1\(\ne\)0

  377. pooja195
    • one year ago
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    \[x \neq 3/2~~~~~x \neq-2~~~x \neq4~~x \neq-1\]

  378. mathmate
    • one year ago
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    So you can write them together as: x\(\ne\)-2,-1,3/2,4

  379. mathmate
    • one year ago
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    Excellent, thank you! :)

  380. pooja195
    • one year ago
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    >:(

  381. mathmate
    • one year ago
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    Ready for another one, simpler, and no conditions!

  382. pooja195
    • one year ago
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    yes!

  383. mathmate
    • one year ago
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    \(\large \frac{4+2x}{x^2-4}.\frac{x^2-4x+4}{x-2}\)

  384. mathmate
    • one year ago
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    Hint: x^2-4 is the same as x^2-2^2, difference of 2 squares.

  385. pooja195
    • one year ago
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    \[\frac{ 4+2x }{ (x+4)(x-4) }\times \frac{ (x-2)(x-2) }{(x-2) }\]

  386. mathmate
    • one year ago
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    and x^2-4x+4 is a perfect square, if you recall!

  387. mathmate
    • one year ago
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    You can do better with the first expression (on the left, both top and bottom.

  388. mathmate
    • one year ago
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    Remember bottom was x^2-2^2

  389. pooja195
    • one year ago
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    2(2+x)

  390. mathmate
    • one year ago
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    good, 2+x= x+2 We usually write polynomials in decreasing power. This way it's easier to find like terms.

  391. mathmate
    • one year ago
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    The bottom is (x+2)(x-2)

  392. pooja195
    • one year ago
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    answer is 1

  393. mathmate
    • one year ago
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    I think it's 2. Can you check?

  394. pooja195
    • one year ago
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    yh its 2 ;o

  395. mathmate
    • one year ago
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    Good. Want to do rational addition/subtraction (more difficult) or rational equations?

  396. pooja195
    • one year ago
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    None of the above. xD

  397. mathmate
    • one year ago
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    ok, it's getting late!

  398. pooja195
    • one year ago
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    LOL ok we can continue tommzzz

  399. pooja195
    • one year ago
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    *tommrow

  400. mathmate
    • one year ago
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    ok!

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