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Math Help - Solving for a variable with fractions

  1. #1
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    Solving for a variable with fractions

    I don't know where to begin on these problems...

    2/3 (x+5) = 1/4 (x+2)

    32(1/16x - 1/32)
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  2. #2
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    Quote Originally Posted by endlesst0m View Post
    I don't know where to begin on these problems...

    2/3 (x+5) = 1/4 (x+2)

    32(1/16x - 1/32)
    2/3 (x+5) = 1/4 (x+2)

    Distribute the fraction

    \frac{2}{3}x+\frac{10}{3}=\frac{1}{4}x+\frac{2}{4}

    Combine the x's on one side and non-x's or numbers on the other side

    \left(\frac{2}{3}-\frac{1}{4}\right)x=\frac{2}{4}-\frac{10}{3}

    Find the lowest common denominators

    \left(\frac{8}{12}-\frac{3}{12}\right)x=\frac{6}{12}-\frac{40}{12}

    \frac{5}{12}x=\frac{-34}{12}

    Multiply by 12

    5x=-34

    Divide by 5

    x = \frac{-34}{5}

    Did that help?
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  3. #3
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    I understand everything except distributing the fractions. What is the procedure and rules for doing that?
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    Quote Originally Posted by endlesst0m View Post
    I understand everything except distributing the fractions. What is the procedure and rules for doing that?
    Hi, endlesst0m! Distribution of multiplication over addition works the same way with fractions as it does with whole numbers. For example:

    If you had 5(x^2-3), you would distribute the 5 to get 5\cdot x^2 - 5\cdot3=5x^2 - 15.

    The process is the same with fractions:

    \frac23(x+5)=\frac14(x+2)

    \Rightarrow\frac23\cdot x+\frac23\cdot5=\frac14\cdot x+\frac14\cdot2

    \Rightarrow\frac{2x}3+\frac{2\cdot5}3=\frac x4 +\frac{1\cdot2}4

    \Rightarrow\frac{2x}3+\frac{10}3=\frac x4+\frac12

    Do you see it now?
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  5. #5
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    I'm pretty sure I do...you multiply the numerator and the number in parenthesis, and place the product over the denominator. Right?
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    Quote Originally Posted by endlesst0m View Post
    I'm pretty sure I do...you multiply the numerator and the number in parenthesis, and place the product over the denominator. Right?
    Yes; otherwise, if you ended up multiplying both the numerator and the denominator by the same number, you would be, in effect, multiplying by one.
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  7. #7
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    So for the other question I posted:

    32(1/16x - 1/32)

    Would the proper distribution be 1/2x and 1? I got these answers by doing 32 divided by 16 and 32 divided by 32.
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    Quote Originally Posted by endlesst0m View Post
    So for the other question I posted:

    32(1/16x - 1/32)

    Would the proper distribution be 1/2x and 1? I got these answers by doing 32 divided by 16 and 32 divided by 32.
    You should get 2x-1, not \frac12x-1. Here is how:

    32\left(\frac1{16}x - \frac1{32}\right)

    = 32\cdot\frac1{16}x - 32\cdot\frac1{32}

    = \frac{32}{16}x - \frac{32}{32}

    =2x - 1
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