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neverforgetvivistee

when you're subtracting rationals, do you make it into a plus sign and make EVERYTHING in the numerator of the second negative, or just make 1 number negative? for examples x-1 4x+2 ---- - ----- x(x-3) x(x-3) would you make it into a plus sign, and then make 4x and 2 negative? -4x-2

  • 2 years ago
  • 2 years ago

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  1. satellite73
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    yes

    • 2 years ago
  2. neverforgetvivistee
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    in the other equation you did how did you get positive 2?

    • 2 years ago
  3. neverforgetvivistee
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    the last one you answered

    • 2 years ago
  4. satellite73
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    \[\frac{x-1-(4x+2)}{x(x-3)}\] \[\frac{x-1-4x-2}{x(x-3)}\] etc

    • 2 years ago
  5. satellite73
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    well in the last one you caught me in a dumb algebra mistake that is why!

    • 2 years ago
  6. satellite73
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    should be \[\frac{x+2}{x}=\frac{x-1}{x}-\frac{4x+2}{x(x-3)}\] \[(x+2)(x-3)=(x-1)(x+3)-4x-2\] \[x^2-x-6 = x^2-2 x-5\] \[-x-6=-2x-5\] \[x=1\]

    • 2 years ago
  7. neverforgetvivistee
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    wait what? i get x^2-x-6=x^2-8x+1 i keep on looking it over but i can't figure it out darn it

    • 2 years ago
  8. satellite73
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    oh man did i mess up again? hold on

    • 2 years ago
  9. neverforgetvivistee
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    ok so, (x-3) and (x+2) equal x^2-x-6. (x-1) and (x-3) euqlas x^2-4x+3. then when i combine them x^2-x-6=x^2-4x+3x-4x-2

    • 2 years ago
  10. satellite73
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    ok now you have caught me in two mistakes, so i am going to quit. you are right it is \[(x+2)(x-3)=(x-1)(x-3)-4x-2\]

    • 2 years ago
  11. neverforgetvivistee
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    then i combine like terms and get x^2-x-6=x^2-8x+1

    • 2 years ago
  12. neverforgetvivistee
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    no please don't this is a very confusing problem lol

    • 2 years ago
  13. neverforgetvivistee
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    then i get 2x^1+7x+7

    • 2 years ago
  14. satellite73
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    ok but i am making mistake after mistake. lets see what we get now

    • 2 years ago
  15. neverforgetvivistee
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    i mean 2x^2+7x+7

    • 2 years ago
  16. satellite73
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    did you get \[x^2-x-6 = x^2-8 x+1\]?

    • 2 years ago
  17. neverforgetvivistee
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    yes

    • 2 years ago
  18. satellite73
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    ok subtract \[x^2\] from both sides

    • 2 years ago
  19. satellite73
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    let me know what you get

    • 2 years ago
  20. neverforgetvivistee
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    7x+7

    • 2 years ago
  21. satellite73
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    what happened to your equal sign?

    • 2 years ago
  22. neverforgetvivistee
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    0=7x+7

    • 2 years ago
  23. neverforgetvivistee
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    oh wait let me do this again

    • 2 years ago
  24. neverforgetvivistee
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    OMFG

    • 2 years ago
  25. satellite73
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    \[x^2-x-6 = x^2-8 x+1\] \[\cancel{x^2}-x-6=\cancel{x^2}-8x+1\] \[-x-6=-8x+1\]

    • 2 years ago
  26. neverforgetvivistee
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    I GOT THE RIGHT ANSWER THANK YOU!!!!!!!!!!!!!! FINALLY THIS PROBLEM IS THE WORST THANK YOU!!!!!!!!!!!!!

    • 2 years ago
  27. neverforgetvivistee
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    YES!!!!!!!!!

    • 2 years ago
  28. neverforgetvivistee
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    took forever!!!!!!!

    • 2 years ago
  29. neverforgetvivistee
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    x=1

    • 2 years ago
  30. satellite73
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    yw. now i have a question. what is vivistee?

    • 2 years ago
  31. neverforgetvivistee
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    my cat that ran away

    • 2 years ago
  32. neverforgetvivistee
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    never forget

    • 2 years ago
  33. neverforgetvivistee
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    listen thanks alright? my gosh that problem was so confusing and frustrating

    • 2 years ago
  34. neverforgetvivistee
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    :"}

    • 2 years ago
  35. neverforgetvivistee
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    it's finally over now :"}

    • 2 years ago
  36. neverforgetvivistee
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    i'm going to write that one down in my notes

    • 2 years ago
  37. satellite73
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    glad you got it, sorry about your cat

    • 2 years ago
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