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anonymous

  • one year ago

(5c-4d)/(2c)-(8c-7d)/(3c)+4 write as a single fraction

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  1. anonymous
    • one year ago
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    \[\frac{ (5c-4d) }{ 2c }-\frac{ (8c-7d) }{ 3c }+4\] is this your expression?

  2. anonymous
    • one year ago
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    yes!

  3. anonymous
    • one year ago
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    You may use the following \[\frac{ a }{ b } \pm \frac{ c }{ d } = \frac{ ad \pm bc }{ bd }\] try it out

  4. anonymous
    • one year ago
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    sorry am i cross multiplying?

  5. anonymous
    • one year ago
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    The \[\pm \] implies it works for either sign

  6. anonymous
    • one year ago
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    Not quite

  7. anonymous
    • one year ago
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    so without much simplifying, its 5c-4d(3c)±2c(8c-7d)/2c(3c)?

  8. anonymous
    • one year ago
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    It's just like adding/ subtracting regular fractions :)

  9. anonymous
    • one year ago
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    So yes you did it right!

  10. anonymous
    • one year ago
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    except you don't have to write p/m I was just stating it would work for either sign

  11. anonymous
    • one year ago
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    oh okay, but wouldnt the denominator be 6c?

  12. anonymous
    • one year ago
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    \[\frac{ 3c(5c-4d) -2c(8c-7d)}{ (2c)(3c) }+4\]

  13. anonymous
    • one year ago
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    Close but you have \[6c^2\]for the denominator

  14. anonymous
    • one year ago
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    opps haha forgot the ^2

  15. anonymous
    • one year ago
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    :)

  16. anonymous
    • one year ago
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    Keep going, you're on the right track

  17. anonymous
    • one year ago
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    so I got 15c^2-12cd-16c^2-14cd/6c^2

  18. anonymous
    • one year ago
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    would I now just simplify it?

  19. anonymous
    • one year ago
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    Yes keep simplifying and watch the signs \[\frac{ 3c(5c-4d) -2c(8c-7d)}{ (2c)(3c) }+4 \implies \frac{ 15c^2-12cd-16c^2 \color{red}+14cd }{ 6c^2 }+4\]

  20. anonymous
    • one year ago
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    so I got -1c^2+2cd/6c^2

  21. anonymous
    • one year ago
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    +4

  22. anonymous
    • one year ago
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    \[\frac{ -c^2+2cd }{ 6c^2 }+4\] looks good! You can do the same thing with the 4 now, \[4 = \frac{ 4 }{ 1 }\]

  23. anonymous
    • one year ago
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    and so i multiply 4 by 6c^2 to get a common denominator?

  24. anonymous
    • one year ago
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    The denominator will stay as 6c^2 you just have to multiply 4 by 6c^2 in the numerator

  25. anonymous
    • one year ago
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    so it'll be 24c^2

  26. anonymous
    • one year ago
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    \[\frac{ -c^2+2cd }{ 6c^2 }+4 \implies \frac{ -c^2+2cd+4(6c^2) }{ 6c^2 } \implies \frac{ -c^2+2cd+24c^2 }{ 6c^2 }\]

  27. anonymous
    • one year ago
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    for final answer I got 23c^2+2cd/6c^2

  28. anonymous
    • one year ago
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    That works but we can simplify it more \[\frac{ 23c^2+2cd }{ 6c^2 } = \frac{ c(23c+2d) }{ 6c^2 } = \frac{ 23c+2d }{ 6c }\] notice we factored out a c.

  29. anonymous
    • one year ago
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    It looks much more clean this way :P

  30. anonymous
    • one year ago
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    does the c cancel in 23c and 6c?

  31. anonymous
    • one year ago
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    Yeah that c got cancelled out so we're left with just 1 c, so here \[\frac{ 23c^2+2cd }{ 6c^2 } = \frac{ \cancel c(23c+2d) }{ 6c^{\cancel 2} } = \frac{ 23c+2d }{ 6c }\]

  32. anonymous
    • one year ago
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    \[\frac{ c }{ c^2 } = \frac{ 1 }{ c }\]

  33. anonymous
    • one year ago
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    ohhh.. that makes it clear, haha my gosh thank you so much!

  34. anonymous
    • one year ago
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    Yw :)

  35. anonymous
    • one year ago
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    Also try that rule I showed you above with regular fractions and you will see it's the exact same thing as adding or subtracting fractions, just a quick little rule. :) So you can even use it for something like \[\frac{ 1 }{ 3 } + \frac{ 4 }{ 5 }\]

  36. anonymous
    • one year ago
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    it was very helpful, much easier to understand compared to what my teacher taught me :)

  37. anonymous
    • one year ago
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    Yeah, sometimes teachers do that...these days they complicate a lot of things which come from simple things as I showed you above ^.^

  38. anonymous
    • one year ago
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    super helpful, thank you so so much!

  39. anonymous
    • one year ago
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    Yw!

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