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Chiomatn93 Group Title

solve for x.

  • 2 years ago
  • 2 years ago

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  1. Chiomatn93 Group Title
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    |dw:1336187220333:dw|

    • 2 years ago
  2. Australopithecus Group Title
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    or you could just go (x-3)(x-7) = (x-5)(x+2) + (x-5)

    • 2 years ago
  3. maheshmeghwal9 Group Title
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    |dw:1336187350332:dw|

    • 2 years ago
  4. Australopithecus Group Title
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    now expand and simplify

    • 2 years ago
  5. maheshmeghwal9 Group Title
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    Now Simplify & solve. Can you do it now?

    • 2 years ago
  6. UnkleRhaukus Group Title
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    \[\frac{x-3}{x-5}=\frac{x+2}{x-7}+1\] \[\frac{x-3}{x-5}-\frac{x+2}{x-7}-1=0\] \[=\frac{(x-7)(x-3)}{(x-7)(x-5)}-\frac{(x-5)(x+2)}{(x-5)(x-7)}-\frac{(x-5)(x-7)}{(x-5)(x-7)}\] \[=\frac{(x-7)(x-3)-(x-5)(x+2)-(x-5)(x-7)}{(x-7)(x-5)}\]

    • 2 years ago
  7. Chiomatn93 Group Title
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    solve for x!!!

    • 2 years ago
  8. Australopithecus Group Title
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    What are you having problems with exactly? do you know how to simplify equations at all?

    • 2 years ago
  9. Australopithecus Group Title
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    (x-3)(x-7) = (x-5)(x+2) + (x-5) = x^(2) - 10x + 21 = x^(2) - 3x - 10 = Im sure you can solve it from here but if you dont know how I got (x-3)(x-7) = (x-5)(x+2) + (x-5) from your original problem then say so

    • 2 years ago
  10. Australopithecus Group Title
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    I have no idea what the other people are doing in this problem :l

    • 2 years ago
  11. sheg Group Title
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    have u understood how to solve this problem???

    • 2 years ago
  12. Australopithecus Group Title
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    x^(2) - 10x + 21 = x^(2) - 3x - 10 = -10x + 3x = -10 - 21 = -7x = -31 x = -31/-7 = x = 31/7

    • 2 years ago
  13. Australopithecus Group Title
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    that is your answer the rest of the people who attempted this problem need a refresher on algebra and solving equations or they just did things the long way

    • 2 years ago
  14. Chiomatn93 Group Title
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    okay thanks so much lol..

    • 2 years ago
  15. Australopithecus Group Title
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    do you have any idea how I solved this? If you want me to show you how I can.

    • 2 years ago
  16. Chiomatn93 Group Title
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    yes please... im not the sharpest knife in the drawer when it comes to mathematics!

    • 2 years ago
  17. Australopithecus Group Title
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    well first you always want to bring all x terms into the numerator so to do this I multiplied both sides of the equation by (x-5) and (x-7) so (x-3)(x-5)(x-7)/(x-5) = (1 + (x+2)/(x-7)) * (x-5)(x-7) = (x-3)(x-7)(1) = (x-5)(x-7) + (x-5)(x-7)(x+2)/(x-7) = (x-3)(x-7) = (x-5)(x-7) + (x-5)(x+2)(1) = (x-5)(x-7) + (x-5)(x-7)(x+2) = (x-5)(x-7) + (x-5)(x+2) = Now expand everything Looks like I made a mistake :S

    • 2 years ago
  18. Australopithecus Group Title
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    in my previous answer

    • 2 years ago
  19. Australopithecus Group Title
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    ugh my brain it is dead. (x-3)(x-5)(x-7)/(x-5) = (1 + (x+2)/(x-7)) * (x-5)(x-7) = (x-3)(x-7)(1) = (x-5)(x-7) + (x-5)(x-7)(x+2)/(x-7) = (x-3)(x-7) = (x-5)(x-7) + (x-5)(x+2)(1) = (x-3)(x-7) = (x-5)(x-7) + (x-5)(x+2) Now expand and simplify to get the answer

    • 2 years ago
  20. Australopithecus Group Title
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    collect all x terms on one side of the equation and move all plain numbers on the other side of the equation

    • 2 years ago
  21. Australopithecus Group Title
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    this problem is quadratic so yeah there are two solutions https://www.wolframalpha.com/input/?i=%28x-3%29%28x-7%29+%3D+%28x-5%29%28x-7%29+%2B+%28x-5%29%28x%2B2%29

    • 2 years ago
  22. Chiomatn93 Group Title
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    alright thanks a bunch!

    • 2 years ago
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