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Math Help - Simple algebra question, solve for the variable

  1. #1
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    Simple algebra question, solve for the variable

    I'm a little rusty here (or possibly losing my mind at this point), but actually this question involves a bigger problem (substitution rule/calculus) but I'm trying to understand a simple algebra concept first. See the following:

    y = 2;
    x = 10;

    x/y = 5

    Ok. I want to solve for y now. Exactly how is this done in algebra? What do you divide or multiply both sides by? I know the answer is y = 1/5x, but I'm unsure as to how one would get to that.

    The only way I could see it is to solve for x first:

    x = 5y, and then divide 5 on both sides: x/5 = y. There must be an easier and non redundant way, correct?
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  2. #2
    Super Member Bacterius's Avatar
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    Uhm, you just said it, y = 2

    But to solve \frac{x}{y} = 5, you can rearrange as follows :

    >> For x :

    \frac{x}{y} = 5

    \frac{x}{y} \times y = 5 \times y

    \frac{x \times y}{y} = 5 \times y

    \frac{x}{1} = 5 \times y (cancel out the y)

    x = 5 \times y

    x = 5y

    >> For y :

    \frac{x}{y} = 5

    \frac{y}{x} = \frac{1}{5} (inverse on both sides)

    \frac{y}{x} \times x = \frac{1}{5} \times x

    \frac{xy}{x} = \frac{x}{5}

    \frac{y}{1} = \frac{x}{5} (cancelling out)

    y = \frac{x}{5}

    y = \frac{1}{5} x

    This is how I would do it (the long way of course, in real situations I would spare most steps). Does it make sense ?
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  3. #3
    MHF Contributor
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    Quote Originally Posted by Bacterius View Post
    Uhm, you just said it, y = 2

    But to solve \frac{x}{y} = 5, you can rearrange as follows :

    >> For x :

    \frac{x}{y} = 5

    \frac{x}{y} \times y = 5 \times y

    \frac{x \times y}{y} = 5 \times y

    \frac{x}{1} = 5 \times y (cancel out the y)

    x = 5 \times y

    x = 5y

    >> For y :

    \frac{x}{y} = 5

    \frac{y}{x} = \frac{1}{5} (inverse on both sides)

    \frac{y}{x} \times x = \frac{1}{5} \times x

    \frac{xy}{x} = \frac{x}{5}

    \frac{y}{1} = \frac{x}{5} (cancelling out)

    y = \frac{x}{5}

    y = \frac{1}{5} x

    This is how I would do it (the long way of course, in real situations I would spare most steps). Does it make sense ?
    You went all out.
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