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Math Help - Solving an equation.

  1. March 18th 2009, 12:10 PM #1
    strunz
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    Solving an equation.

    How about solving this:

    6/5+3/(w-3)=9/[5(w-3)]

    LCD= 5(w-3)

    (6/5)(5)(w-3)+5(w-3)[3/(w-3)]=9/[5(w-3)](5)(w-3)

    6w-18+5w-15(3/(w-3)=9/[5(w-3)](5)(w-3)


    5w-15(3/(w-3)=9/[5(w-3)](5)(w-3) that's the part I'm having trouble with..
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  2. March 18th 2009, 02:13 PM #2
    stapel
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    Quote Originally Posted by strunz View Post
    How about solving this:

    6/5+3/(w-3)=9/[5(w-3)]
    You appear to have:

    . . . . . \frac{6}{5}\, +\, \frac{3}{w\, -\, 3}\, =\, \frac{9}{5(w\, -\, 3)}

    Multiplying through by 5(w - 3) gives:

    . . . . . 6(w\, -\, 3)\, +\, 3(5)\, =\, 9

    I'm not sure how you got your result...?
    Last edited by mr fantastic; March 21st 2009 at 02:20 PM. Reason: Edit by stapel: Correcting typo. Edit by Mr F: Moved posts.
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  3. March 19th 2009, 01:45 AM #3
    Showcase_22
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    <br /> \frac{6}{5}\, +\, \frac{3}{w\, -\, 3}\, =\, \frac{9}{5(w\, -\, 3)}<br />

    6(w-3)+3(5)=9? (multiplying through by 5(w-3)).

    Or is my mental arithmetic terrible?

    6w-18+15=9

    6w-3=9

    6w=12

    w=2.
    Last edited by Showcase_22; March 19th 2009 at 01:46 AM. Reason: typo!
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  4. March 19th 2009, 04:17 AM #4
    stapel
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    Quote Originally Posted by Showcase_22 View Post
    w=2.
    Check your solution by plugging it back in to the original equation:

    . . . . . \mbox{left-hand side: }\, \frac{6}{5}\, +\, \frac{3}{(2)\, -\, 3}\, =\, \frac{6}{5}\, +\, \frac{3}{-1}\, =\, \frac{6}{5}\, -\, \frac{15}{5}\, =\, -\frac{9}{5}

    . . . . . \mbox{right-hand side: }\, \frac{9}{5\left[(2)\, -\, 3\right]}\, =\, \frac{9}{5(-1)}\, =\, -\frac{9}{5}

    Looks good to me!
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  5. March 19th 2009, 11:34 AM #5
    strunz
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    Fractional equation w/ binomial denominator

    <br /> \frac{6}{5}\, +\, \frac{3}{w\, -\, 3}\, =\, \frac{9}{5(w\, -\, 3)}<br />

    Hi could someone write down in steps how to actually solve this.. thanks
    Last edited by mr fantastic; March 21st 2009 at 02:21 PM. Reason: Part of the mass movement
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  6. March 19th 2009, 11:43 AM #6
    e^(i*pi)
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    Quote Originally Posted by strunz View Post
    <br /> \frac{6}{5}\, +\, \frac{3}{w\, -\, 3}\, =\, \frac{9}{5(w\, -\, 3)}<br />

    Hi could someone write down in steps how to actually solve this.. I already have the answer but can't figure out how the steps precede.. thanks
    get the same denominator of 5(w-3)

    \frac{6}{5} \cdot \frac{w-3}{w-3} + \frac{3}{w-3} \cdot \frac{5}{5} - \frac{9}{5(w-3)} = 0

    now you can collect all the terms

    6w-18 + 15 + 9 = 0
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  7. March 19th 2009, 01:09 PM #7
    strunz
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    <br /> <br /> \frac{6}{5}\, +\, \frac{3}{w\, -\, 3}\, =\, \frac{9}{5(w\, -\, 3)}<br />

    My book says the second step precedes like this

    (6/5)(5)(w-3)(3/(w-3)=9/[5(w-3)](5)(w-3)

    6(w-3)+15=9
    w=2

    Maybe I'm not at the level you think I am, because my problem is really how to multiplicate those parentheses.. the bold written line.. I don't know the way it's done.
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  8. March 20th 2009, 04:02 PM #8
    fxsapa
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    Ok, first write that properly:

    First line:
    <br /> <br /> \frac{6}{5}\, +\, \frac{3}{w\, -\, 3}\, =\, \frac{9}{5(w\, -\, 3)}<br />

    Second line:
    <br /> <br /> \frac{6 \times 5(w-3)}{5} + \frac{3\times 5(w-3)}{w-3} = \frac{9}{5(w-3)} \times 5(w-3)<br /> <br />

    What has been done between lines 1 & 2:

    To get 2, you simply multiply both sides by 5(w-3). This is one way of solving an equation with fractions.

    From there, you need to simplify the expression. Cross out terms... and get rid of the fractions:

    <br /> <br /> 6 \times (w-3) + 3 \times 5 = 9<br /> <br />


    That done, multiply the terms and finish to solve!
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  9. March 20th 2009, 04:50 PM #9
    e^(i*pi)
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    Quote Originally Posted by strunz View Post
    <br /> <br /> \frac{6}{5}\, +\, \frac{3}{w\, -\, 3}\, =\, \frac{9}{5(w\, -\, 3)}<br />

    My book says the second step precedes like this

    (6/5)(5)(w-3)(3/(w-3)=9/[5(w-3)](5)(w-3)

    6(w-3)+15=9
    w=2

    Maybe I'm not at the level you think I am, because my problem is really how to multiplicate those parentheses.. the bold written line.. I don't know the way it's done.
    You have the same answer I do, I just used the lowest common denominator which is 5(w-3)
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  10. March 21st 2009, 04:16 AM #10
    HallsofIvy
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    The simplest way to do this is not to combine fractions but to get rid of the fractions by multiplying on both sides of the equation by the common denominator 5(w-3):

    5(w-3)\left[\frac{6}{5}+ \frac{3}{w-3}\right]= 5(w-3)\left[\frac{9}{5(w- 3)}\right]
    6(w-3)+ 15= 9
    immediately.
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  11. March 21st 2009, 06:42 AM #11
    stapel
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    Quote Originally Posted by HallsofIvy View Post
    The simplest way to do this is not to combine fractions but to get rid of the fractions by multiplying on both sides of the equation by the common denominator 5(w-3):

    5(w-3)\left[\frac{6}{5}+ \frac{3}{w-3}\right]= 5(w-3)\left[\frac{9}{5(w- 3)}\right]
    6(w-3)+ 15= 9
    immediately.
    That's what the poster was shown earlier in the thread. I'm guessing maybe the poster isn't allowed to skip any steps at all...?
    Last edited by mr fantastic; March 21st 2009 at 02:22 PM. Reason: Part of the mass movement.
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  12. March 21st 2009, 11:23 AM #12
    HallsofIvy
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    I wouldn't consider that "skipping" any steps at all.
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