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vera_ewing

  • one year ago

How do you solve this?

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  1. vera_ewing
    • one year ago
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  2. jim_thompson5910
    • one year ago
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    They provided the definition of \(\Large \boxed{x}\) \[\Large \boxed{x} = \frac{x+3}{x-1}\] subtract 1 from both sides to get \[\Large \boxed{x}-1 = \frac{x+3}{x-1}-1\] now simplify the right hand side

  3. vera_ewing
    • one year ago
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    Wait why do I subtract 1 from both sides?!

  4. jim_thompson5910
    • one year ago
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    because they want to know what \(\large \boxed{x}-1\) is equivalent to

  5. vera_ewing
    • one year ago
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    So we just put a -1 on both sides? That makes no sense...

  6. jim_thompson5910
    • one year ago
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    you start off with \(\large \boxed{x}\) and do that operation to get \(\large \boxed{x}-1\)

  7. vera_ewing
    • one year ago
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    But there is no -1 originally. Where did it come from?

  8. jim_thompson5910
    • one year ago
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    maybe it might make more sense if y = 2x was given to you if that's the case, then you would subtract 1 from both sides to get y-1 = 2x-1

  9. jim_thompson5910
    • one year ago
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    I'm doing that operation of subtract 1 from both sides to turn \(\large \boxed{x}\) into \(\large \boxed{x}-1\) basically it's like turning y into y-1

  10. vera_ewing
    • one year ago
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    I'm so sorry, but why are we allowed to subtract 1 from both sides? There is no 1 on either side of the equation....

  11. jim_thompson5910
    • one year ago
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    This is what the full step by step picture looks like \[\Large \boxed{x} = \frac{x+3}{x-1}\] \[\Large \boxed{x}-1 = \frac{x+3}{x-1}-1\] \[\Large \boxed{x}-1 = \frac{x+3}{x-1}-1*\frac{x-1}{x-1}\] \[\Large \boxed{x}-1 = \frac{x+3}{x-1}-\frac{x-1}{x-1}\] \[\Large \boxed{x}-1 = \frac{x+3-(x-1)}{x-1}\] \[\Large \boxed{x}-1 = \frac{x+3-x+1}{x-1}\] \[\Large \boxed{x}-1 = \frac{4}{x-1}\]

  12. jim_thompson5910
    • one year ago
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    I'm using the subtraction property of equality if a = b, then a-c = b-c

  13. vera_ewing
    • one year ago
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    Ohh okay. And why is there a box around the x on the left?

  14. jim_thompson5910
    • one year ago
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    because that's how they set up the notation in the problem

  15. vera_ewing
    • one year ago
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    Does it mean anything?

  16. jim_thompson5910
    • one year ago
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    it's just their fancy way of defining a function

  17. jim_thompson5910
    • one year ago
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    they could have easily said "let y = ..." or "let f(x) = ..."

  18. jim_thompson5910
    • one year ago
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    why they chose a box, who knows

  19. vera_ewing
    • one year ago
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    Okay. Can you please give me a similar problem to solve?

  20. vera_ewing
    • one year ago
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    I need to be able to do these on my own.

  21. jim_thompson5910
    • one year ago
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    ok what is \[\Large \boxed{x} +2x\] equivalent to? (same function from the previous problem)

  22. vera_ewing
    • one year ago
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    I'm not sure :(

  23. jim_thompson5910
    • one year ago
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    replace the box x with what it is equal to then simplify

  24. vera_ewing
    • one year ago
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    4/x-1 ?

  25. jim_thompson5910
    • one year ago
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    no

  26. jim_thompson5910
    • one year ago
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    box x was equal to (x+3)/(x-1)

  27. jim_thompson5910
    • one year ago
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    so saying \[\Large \boxed{x}+2x\]is the same as saying (x+3)/(x-1) + 2x. Simplify from here

  28. vera_ewing
    • one year ago
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    What would the first step be?

  29. jim_thompson5910
    • one year ago
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    you need to get the denominators the same

  30. jim_thompson5910
    • one year ago
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    2x = 2x/1 has a denominator of 1

  31. jim_thompson5910
    • one year ago
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    how can you transform that 1 into (x-1)

  32. vera_ewing
    • one year ago
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    So (x+3)/(x-1) + 2x/1 ?

  33. jim_thompson5910
    • one year ago
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    how can you get 2x/1 to have a denominator of x-1

  34. vera_ewing
    • one year ago
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    Multiply 1 on both sides?

  35. jim_thompson5910
    • one year ago
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    why not multiply top and bottom by (x-1) ?

  36. jim_thompson5910
    • one year ago
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    why not multiply top and bottom of 2x/1 by (x-1) ?

  37. jim_thompson5910
    • one year ago
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    you might be mixing up + and =

  38. vera_ewing
    • one year ago
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    Show me what that step would look like please.

  39. jim_thompson5910
    • one year ago
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    |dw:1434499487201:dw|

  40. jim_thompson5910
    • one year ago
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    |dw:1434499531547:dw|

  41. jim_thompson5910
    • one year ago
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    |dw:1434499550135:dw|

  42. jim_thompson5910
    • one year ago
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    |dw:1434499580374:dw|

  43. jim_thompson5910
    • one year ago
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    I'll let you finish

  44. vera_ewing
    • one year ago
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    Hold on, this all makes so much sense, but I just don't understand what property lets us multiply (x-1) on the top and bottom.

  45. jim_thompson5910
    • one year ago
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    I'm essentially multiplying by 1 (x-1)/(x-1) = 1

  46. jim_thompson5910
    • one year ago
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    |dw:1434499775976:dw|