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18jonea

  • one year ago

Add/Subtract: : Simplify and state the domain.

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  1. 18jonea
    • one year ago
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    |dw:1440004351229:dw|

  2. 18jonea
    • one year ago
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    @Michele_Laino

  3. freckles
    • one year ago
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    can't really understand this? what is going on with that 1/4 over there... And I see you have - + in between the first two fractions.

  4. 18jonea
    • one year ago
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    the plus is suppose to be over before the 1/4

  5. freckles
    • one year ago
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    \[\frac{7}{x+6}-\frac{x+2}{2x+2}+\frac{1}{4}?\]

  6. 18jonea
    • one year ago
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    yes

  7. freckles
    • one year ago
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    well first 2x+2 can be written as 2(x+1) \[\frac{7}{x+6}-\frac{x+2}{2(x+1)}+\frac{1}{4}\] now look at the bottoms of all the fractions and ask yourself how you can find a common denominator

  8. 18jonea
    • one year ago
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    no?

  9. freckles
    • one year ago
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    what does that mean exactly

  10. freckles
    • one year ago
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    "no?" I don't understand the response to what I said.

  11. freckles
    • one year ago
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    does that mean you don't know how to find a common denominator ?

  12. 18jonea
    • one year ago
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    i dont know how to find the common denominator

  13. freckles
    • one year ago
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    oh okay

  14. freckles
    • one year ago
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    well if you notice the denominators are x+6 and 2(x+1) and 4

  15. freckles
    • one year ago
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    so a common denominator could be (x+6)*(x+1)*4

  16. freckles
    • one year ago
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    the first fraction... \[\frac{7}{x+6}\] how can we write this as something over/[4(x+6)(x+1)]?

  17. 18jonea
    • one year ago
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    {4(x+6)(x+1)}

  18. freckles
    • one year ago
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    notice on bottom it is lacking the 4 and the (x+1) so just multiply both top and bottom by 4(x+1) that is how you can get this new bottom we speak of ... \[\frac{7}{x+6} \\ \text{ multiply top and bottom by } 4(x+1) \\ \frac{7}{x+6} \cdot \frac{4(x+1)}{4(x+1)}=\frac{7 \cdot 4(x+1)}{4(x+6)(x+1)}\]

  19. freckles
    • one year ago
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    now look at the second fraction

  20. freckles
    • one year ago
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    \[-\frac{x+2}{2(x+1)}\] remember we want 4(x+6)(x+1) on bottom so what is this bottom lacking in order to get that?

  21. 18jonea
    • one year ago
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    x+6

  22. freckles
    • one year ago
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    and also 2 right since 2*2=4

  23. 18jonea
    • one year ago
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    i am so confused

  24. freckles
    • one year ago
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    \[-\frac{x+2}{2(x+1)} \cdot \frac{2(x+6)}{2(x+6)} =-\frac{2(x+2)(x+6)}{2 \cdot 2 (x+6)(x+1)}\]

  25. freckles
    • one year ago
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    we wanted 4(x+1)(x+6) on bottom

  26. freckles
    • one year ago
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    we had 2(x+1) on bottom

  27. freckles
    • one year ago
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    2(x+1)*what=4(x+1)(x+6)

  28. freckles
    • one year ago
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    divide both sides 2(x+1) \[what=\frac{4(x+1)(x+6)}{2(x+1)}=\frac{4}{2} \frac{(x+1)}{(x+1)} \frac{x+6}{1} \\ what=2 (1)(x+6) \\ what=2(x+6)\] so 2(x+1)*2(x+6)=4(x+1)(x+6)

  29. freckles
    • one year ago
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    so what we needed to do to get the bottom 4(x+1)(x+6) for the second fraction was multiply the second fraction on top and bottom by 2(x+6)

  30. freckles
    • one year ago
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    so far we have: \[\frac{7}{x+6}-\frac{x+2}{2(x+1)}+\frac{1}{4} \\ = \\ \frac{7 \cdot 4(x+1)}{4(x+6)(x+1)}-\frac{2(x+6)(x+2)}{4(x+1)(x+6)}+\frac{1}{4} \\ \text{ so we are already able to combine the first two fractions } \\ \frac{7 \cdot 4(x+1)-2(x+6)(x+2)}{4(x+1)(x+6)}+\frac{1}{4}\]

  31. freckles
    • one year ago
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    you still need to do the same thing you did to the first two fractions to the third

  32. freckles
    • one year ago
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    what do we need to multiply 4 by so that we have 4(x+1)(x+6)

  33. 18jonea
    • one year ago
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    x+1 x+6

  34. freckles
    • one year ago
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    right so multiply top and bottom by (x+1)(x+6)

  35. freckles
    • one year ago
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    \[\frac{1}{4}=\frac{1(x+1)(x+6)}{4(x+1)(x+6)} \text{ or just } =\frac{(x+1)(x+6)}{4(x+1)(x+6)}\]

  36. freckles
    • one year ago
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    so this gives us: \[\frac{7}{x+6}-\frac{x+2}{2(x+1)}+\frac{1}{4} \\ = \\ \frac{7 \cdot 4(x+1)}{4(x+6)(x+1)}-\frac{2(x+6)(x+2)}{4(x+1)(x+6)}+\frac{1}{4} \\ \text{ so we are already able to combine the first two fractions } \\ \frac{7 \cdot 4(x+1)-2(x+6)(x+2)}{4(x+1)(x+6)}+\frac{1}{4}\] \[\frac{7 \cdot 4(x+1)-2(x+6)(x+2)+(x+1)(x+6)}{4(x+1)(x+6)}\]

  37. freckles
    • one year ago
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    you should multiply everything out on top and combine any like terms

  38. freckles
    • one year ago
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    also notice the domain can be found from the start

  39. freckles
    • one year ago
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    you don't want any of the fractions to be 0 on bottom so to find the domain say all real numbers except x+1=0 or x+6=0 (I will let you solve those equations)

  40. 18jonea
    • one year ago
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    why is there a x+2

  41. 18jonea
    • one year ago
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    and x+1

  42. freckles
    • one year ago
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    the second fraction had x+2 in the numerator

  43. freckles
    • one year ago
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    and the second fraction also had 2x+2 on bottom 2x+2=2(x+1)

  44. 18jonea
    • one year ago
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    right

  45. 18jonea
    • one year ago
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    is the domain all real numbers except when x = 6 or 1

  46. 18jonea
    • one year ago
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    @freckles

  47. 18jonea
    • one year ago
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    wait it would be -6 right?

  48. freckles
    • one year ago
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    -6 is right 1 is incorrect this is because 1+1=0 isn't a true equation

  49. freckles
    • one year ago
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    -1+1=0 is true

  50. freckles
    • one year ago
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    x+1=0 means x=-1

  51. freckles
    • one year ago
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    the domain is all real numbers except x=-1 or x=-6

  52. 18jonea
    • one year ago
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    right i just figured that out

  53. freckles
    • one year ago
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    did you ever try multiplying the top and combining like terms yet?

  54. 18jonea
    • one year ago
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    i am trying to figure out the second equation you did... can you show me how you did it again please

  55. freckles
    • one year ago
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    fraction?

  56. 18jonea
    • one year ago
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    yes the second fraction

  57. freckles
    • one year ago
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    this was the second fraction \[-\frac{x+2}{2(x+1)}\]

  58. freckles
    • one year ago
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    we wanted the bottom to be 4(x+1)(x+6)

  59. freckles
    • one year ago
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    the bottom is missing factors 2 and (x+6)

  60. freckles
    • one year ago
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    so we multiply top and bottom by 2(x+6)

  61. 18jonea
    • one year ago
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    so how would it look

  62. freckles
    • one year ago
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    \[-\frac{x+2}{2(x+1)} \cdot \frac{2(x+6)}{2(x+6)} \\ =-\frac{2(x+2)(x+6)}{2 \cdot 2(x+1)(x+6)} \\ = -\frac{2(x+2)(x+6)}{4(x+1)(x+6)}\]

  63. 18jonea
    • one year ago
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    ok i am goimg to try and multiply now

  64. freckles
    • one year ago
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    ok

  65. 18jonea
    • one year ago
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    yeah i have no idea how to even start... can you help

  66. 18jonea
    • one year ago
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    @freckles

  67. 18jonea
    • one year ago
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    wouldnt the second one be 2x squared +16x + 24

  68. 18jonea
    • one year ago
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    @freckles

  69. freckles
    • one year ago
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    2(x+2)(x+6) = 2(x^2+6x+2x+12) 2(x^2+8x+12) 2x^2+16x+24 is right and then there is a negative in front of that mess -2(x+2)(x+6) so it is actually -2x^2-16x-24

  70. freckles
    • one year ago
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    \[\frac{7 \cdot 4(x+1)-2(x+6)(x+2)+(x+1)(x+6)}{4(x+1)(x+6)} \\ \frac{28(x+1)-2x^2-16x-24+(x+1)(x+6)}{4(x+1)(x+6)}\]

  71. freckles
    • one year ago
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    28(x+1)=28x+28 and (x+1)(x+6)=x^2+6x+1x+6 (x+1)(x+6)+x^2+7x+6

  72. freckles
    • one year ago
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    \[\frac{28x+28-2x^2-16x-24+x^2+7x+6}{4(x+1)(x+6)}\]

  73. freckles
    • one year ago
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    combine like terms on top

  74. 18jonea
    • one year ago
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    to combine all of them i got -x^2 +12x +10 All divided by 4(x+6)(x+1)

  75. 18jonea
    • one year ago
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    is that right?

  76. freckles
    • one year ago
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    \[28x+28-2x^2-16x-24+x^2+7x+6 \\ \text{ reorder terms } \\ (-2x^2+x^2)+(28x-16x+7x)+(28-24+6) \\ (-2+1)x^2+(28-16+7)x+(28-24+6) \\ -1x^2+19x+10 \\ -x^2+19x+10\] is what I got for the numerator you might want to check my work

  77. 18jonea
    • one year ago
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    yes thats what i got

  78. 18jonea
    • one year ago
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    sorry i meanyt 19 x

  79. 18jonea
    • one year ago
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    then how would i siplify it

  80. freckles
    • one year ago
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    in my mind \[\frac{-x^2+19x+10}{4(x+1)(x+6)}\] this is simplified

  81. freckles
    • one year ago
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    however your teacher may want you to multiply the bottom out

  82. freckles
    • one year ago
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    not sure in your teacher's mind

  83. 18jonea
    • one year ago
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    thank you

  84. freckles
    • one year ago
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    np

  85. 18jonea
    • one year ago
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    can you help me with one more of these problems that similar..

  86. freckles
    • one year ago
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    you can show me your work and I can help you fix it if it needs any fixing

  87. 18jonea
    • one year ago
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    ok i will

  88. 18jonea
    • one year ago
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    \[\frac{ 1 }{ 2} +\frac{ 15 }{ 2x-14 } - \frac{ 3x-5 }{ x^2-7x }\]

  89. 18jonea
    • one year ago
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    first one stays the same 2(x-7) x(x-7)

  90. freckles
    • one year ago
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    check mark so far! :)

  91. 18jonea
    • one year ago
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    domain is all real numbers except when x=7

  92. freckles
    • one year ago
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    well or x=0

  93. freckles
    • one year ago
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    you have x on the bottom

  94. freckles
    • one year ago
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    on that 3rd fraction

  95. freckles
    • one year ago
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    \[\frac{1}{2}+\frac{15}{2(x-7)}+\frac{3x-5}{x(x-7)} \\ \text{ so we want } x-7 \neq 0 \text{ and also we want } x \neq 0\]

  96. 18jonea
    • one year ago
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    ok

  97. freckles
    • one year ago
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    do you want to try to figure out what the denominators should be so we can combine the fractions? or do you want me to tell you and you try to figure out the new numerators?

  98. 18jonea
    • one year ago
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    you tell me and i try to figure out the numerators

  99. freckles
    • one year ago
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    ok notice the denominators are: 2 2(x-7) x(x-7) ---- if we look at just 2 and 2(x-7) then the denominator would be 2(x-7) but we also have x(x-7) for the third bottom so looking at 2(x-7) and x(x-7) the new denominator of all them should be 2x(x-7) this is because all the denominators given go into 2x(x-7) that I can think of something to multiply all of your original denominators to get this new denominator of 2x(x-7) for example: 2*x(x-7) is 2x(x-7) and 2(x-7)*x is 2x(x-7) and x(x-7)*2 is 2x(x-7)

  100. 18jonea
    • one year ago
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    \[\frac{ 1 }{ 2 } * x(x-7)\]

  101. 18jonea
    • one year ago
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    x * x(x-7)

  102. 18jonea
    • one year ago
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    is that right

  103. freckles
    • one year ago
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    well you multiply top and bottom by x(x-7)

  104. freckles
    • one year ago
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    \[\frac{1}{2} \cdot \frac{x(x-7)}{x(x-7)}\]

  105. freckles
    • one year ago
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    \[\frac{x(x-7)}{2x(x-7)}\] but yep this right for the first fraction

  106. 18jonea
    • one year ago
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    ok

  107. 18jonea
    • one year ago
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    the next one you only need to multiply top and bottom by x

  108. 18jonea
    • one year ago
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    and the last one you only multiply it by 2

  109. freckles
    • one year ago
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    (well 2/2 but I get what you are saying) right!!! so we have: \[\frac{1}{2}+\frac{15}{2(x-7)}+\frac{3x-5}{x(x-7)} \\ =\frac{x(x-7)}{2x(x-7)}+\frac{15x}{2x(x-7)}+\frac{2(3x-5)}{2x(x-7)}\]

  110. 18jonea
    • one year ago
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    thats what i got

  111. freckles
    • one year ago
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    so now we have the same denominators we can now combine the fractions

  112. freckles
    • one year ago
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    \[\frac{1}{2}+\frac{15}{2(x-7)}+\frac{3x-5}{x(x-7)} \\ =\frac{x(x-7)}{2x(x-7)}+\frac{15x}{2x(x-7)}+\frac{2(3x-5)}{2x(x-7)} \\ =\frac{x(x-7)+15x+2(3x-5)}{2x(x-7)}\]

  113. freckles
    • one year ago
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    multiply top out and combine like terms

  114. 18jonea
    • one year ago
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    x^2 +2x - 10

  115. 18jonea
    • one year ago
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    so i can take the 2x out of the top and bottom and then i would get \[\frac{ x^2-10 }{ x-7 }\]

  116. 18jonea
    • one year ago
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    @freckles

  117. freckles
    • one year ago
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    sorry my internet stopped working for a few minutes there

  118. freckles
    • one year ago
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    and you can't cancel out common terms but you can cancel out common factors

  119. freckles
    • one year ago
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    so you should have: \[\frac{1}{2}+\frac{15}{2(x-7)}+\frac{3x-5}{x(x-7)} \\ =\frac{x(x-7)}{2x(x-7)}+\frac{15x}{2x(x-7)}+\frac{2(3x-5)}{2x(x-7)} \\ =\frac{x(x-7)+15x+2(3x-5)}{2x(x-7)} \\ =\frac{x^2+14x-20}{2x(x-7)}\]

  120. freckles
    • one year ago
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    @18jonea

  121. 18jonea
    • one year ago
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    so what would be the final equation

  122. 18jonea
    • one year ago
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    @freckles

  123. freckles
    • one year ago
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    \[\frac{x^2+14x-20}{2x(x-7)}\] and you can multiply to the bottom out if you want or if your teacher wants but to me this is simplified

  124. 18jonea
    • one year ago
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    how did you get 14x and -20

  125. 18jonea
    • one year ago
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    the last fraction you are supppose to subtract

  126. freckles
    • one year ago
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    the problem I have is: \[\frac{1}{2}+\frac{15}{2(x-7)}+\frac{3x-5}{x(x-7)}\]

  127. freckles
    • one year ago
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    is that not the right problem?

  128. 18jonea
    • one year ago
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    sorry it was subtract the last fraction

  129. 18jonea
    • one year ago
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    thats how i got \[\frac{ x^2 +2x-10 }{ 2x(x-7) }\]

  130. freckles
    • one year ago
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    \[\frac{1}{2}+\frac{15}{2(x-7)}-\frac{3x-5}{x(x-7)} \\ =\frac{x(x-7)}{2x(x-7)}+\frac{15x}{2x(x-7)}-\frac{2(3x-5)}{2x(x-7)} \\ =\frac{x(x-7)+15x-2(3x-5)}{2x(x-7)} \\ =\frac{x^2-7x+15x-6x+10}{2x(x-7)}\] \[=\frac{x^2+2x+10}{2x(x-7)}\]

  131. 18jonea
    • one year ago
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    wait it is -10

  132. freckles
    • one year ago
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    -2(-5)=10

  133. 18jonea
    • one year ago
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    why is it -2 for the rest it was +2

  134. freckles
    • one year ago
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    you have -2(3x-5) distributive property says a(b+c)=ab+ac so this means -2(3x-5) is -2(3x)+(-2)(-5) or -6x+10

  135. 18jonea
    • one year ago
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    why is it -2 cause it has +2 at bottom

  136. 18jonea
    • one year ago
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    i have 6x-10

  137. 18jonea
    • one year ago
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    hence how i got x2+2x−10 / 2x(x−7)

  138. freckles
    • one year ago
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    \[\frac{1}{2}+\frac{15}{2(x-7)}\color{red}-\frac{3x-5}{x(x-7)} \\ =\frac{x(x-7)}{2x(x-7)}+\frac{15x}{2x(x-7)}\color{red}-\frac{2(3x-5)}{2x(x-7)} \\ =\frac{x(x-7)+15x\color{red}-2(3x-5)}{2x(x-7)} \\ =\frac{x^2-7x+15x-6x+10}{2x(x-7)} \\ =\frac{x^2+2x+10}{2x(x-7)}\]

  139. freckles
    • one year ago
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    do you see that red subtraction symbol I just brought it down

  140. 18jonea
    • one year ago
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    got it thank you!

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