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Math Help - I think the idea is to factor out the x's

  1. #1
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    I think the idea is to factor out the x's

    I have to equations;

    x - 2 / 3, and 2x / 7

    I need to solve for x?

    First I multiply the denomintors;

    3(x - 2) = 7(2x)

    next I think I need to factor out the x's?

    x(x - 2) from here I am lost?

    I am sure I require to get (x......)(x......) but unsure at this point?
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: I think the idea is to factor out the x's

    Do you mean:
    \frac{x-2}{3}=\frac{2x}{7}?
    Multiply both sides with 21 which is the LCM of 3 and 7.

    Note:
    An equation looks like ...=..., so just \frac{2x}{7} is not an equation.
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  3. #3
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    Re: I think the idea is to factor out the x's

    Quote Originally Posted by Siron View Post
    Do you mean:
    \frac{x-2}{3}=\frac{2x}{7}?
    Multiply both sides with 21 which is the LCM of 3 and 7.

    Note:
    An equation looks like ...=..., so just \frac{2x}{7} is not an equation.
    I must be doing something wrong because when I check the solutions they don't match?

    x - 2 = 2x
    3 7

    27(x - 2) = 27(2x)

    27x - 54 = 54x

    27x - 54x = 54

    -27x = 54
    -27 -27

    x = - 2

    -2 -2 = - 1.33
    3

    2 (-2) = - 4
    7 7

    x = - 0.57???

    Something is up I know I can find problems, but I also know I can't find correct solutions?
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: I think the idea is to factor out the x's

    If you mutliply both sides with 21 then you get:
    \frac{21}{3}(x-2)=\frac{21}{7}(2x)
    \Leftrightarrow 7(x-2)=3(2x)
    \Leftrightarrow 7x-14=6x
    \Leftrightarrow 14=x
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  5. #5
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    Re: I think the idea is to factor out the x's

    Quote Originally Posted by Siron View Post
    If you mutliply both sides with 21 then you get:
    \frac{21}{3}(x-2)=\frac{21}{7}(2x)
    \Leftrightarrow 7(x-2)=3(2x)
    \Leftrightarrow 7x-14=6x
    \Leftrightarrow 14=x
    Thanks Siron,

    I would never had thought about inverting the denominators and numerators to do that?
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  6. #6
    MHF Contributor Siron's Avatar
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    Re: I think the idea is to factor out the x's

    You're welcome!
    If you have a polynomial equation with fractions in it then it's useful to multiply each side with the LCM of the denominators of the fractions.
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