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Math Help - Gr11 Fractions

  1. #1
    TH1
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    Gr11 Fractions

    For these fractions all I knew was to find the lowest common denominator and after that I wasnt sure what to do

    3k/2 - K+3/3 = 8 - k+2/4

    3p/4 + p-5/3 = 1/2

    u-3/4 -2 = 3u/2 = 2u+1/5

    U-5/2 = 2u+5
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by TH1 View Post
    3k/2 - K+3/3 = 8 - k+2/4
    Try to use parentheses, 'cause it becomes ambiguous.

    I assume you mean \frac32k-\frac{k+3}3=8-\frac{k+2}4

    Multiply both sides by 12, this yields

    6\cdot3k-4(k+3)=96-3(k+2)

    which is easy to take.

    The other ones have the same treatment.
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  3. #3
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    Hello, TH1;70157!

    Krizalid is absolutely correct . . .

    With an equation, we can eliminate the fractions (legally).


    Your second problem: . \frac{3k}{2}  - \frac{k+3}{3} \:= \:8 - \frac{k+2}{4}

    The LCD is 12. .Multiply through by 12.

    Multiply everything by 12: . 12\cdot\frac{3k}{2} \,- \,12\cdot\frac{k+3}{3} \;\;=\;\;12\cdot8 \,- \,12\cdot\frac{k+2}{4}

    . . and reduce: . \not{1}\!\!\!\not{2}^6\cdot\frac{3k}{\not{2}} \,- \,\not{1}\!\!\!\not{2}^4\cdot\frac{k+3}{\not{3}} \;\;=\;\;12\cdot8 \,- \,\not{1}\!\!\!\not{2}^3\cdot\frac{k+2}{\not{4}}

    . . and we have: . 6(3k) - 4(k+3) \;=\;96 - 3(k+2)


    Can you finish it now?

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