Results 1 to 12 of 12

Thread: Solving an equation.

  1. March 18th 2009, 12:10 PM #1
    strunz
    strunz is offline
    Newbie
    Joined
    Mar 2009
    Posts
    22

    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..
    Follow Math Help Forum on Facebook and Google+
  2. March 18th 2009, 02:13 PM #2
    stapel
    stapel is offline
    MHF Contributor
    Joined
    Mar 2007
    Posts
    1,240

    Talking

    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.
    Follow Math Help Forum on Facebook and Google+
  3. March 19th 2009, 01:45 AM #3
    Showcase_22
    Showcase_22 is offline
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    <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!
    Follow Math Help Forum on Facebook and Google+
  4. March 19th 2009, 04:17 AM #4
    stapel
    stapel is offline
    MHF Contributor
    Joined
    Mar 2007
    Posts
    1,240

    Talking

    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!
    Follow Math Help Forum on Facebook and Google+
  5. March 19th 2009, 11:34 AM #5
    strunz
    strunz is offline
    Newbie
    Joined
    Mar 2009
    Posts
    22

    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
    Follow Math Help Forum on Facebook and Google+
  6. March 19th 2009, 11:43 AM #6
    e^(i*pi)
    e^(i*pi) is offline
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    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
    Follow Math Help Forum on Facebook and Google+
  7. March 19th 2009, 01:09 PM #7
    strunz
    strunz is offline
    Newbie
    Joined
    Mar 2009
    Posts
    22
    <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.
    Follow Math Help Forum on Facebook and Google+
  8. March 20th 2009, 04:02 PM #8
    fxsapa
    fxsapa is offline
    Newbie
    Joined
    Mar 2009
    Posts
    3
    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!
    Follow Math Help Forum on Facebook and Google+
  9. March 20th 2009, 04:50 PM #9
    e^(i*pi)
    e^(i*pi) is offline
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    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)
    Follow Math Help Forum on Facebook and Google+
  10. March 21st 2009, 04:16 AM #10
    HallsofIvy
    HallsofIvy is offline
    MHF Contributor
    Joined
    Apr 2005
    Posts
    13,727
    Thanks
    550
    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.
    Follow Math Help Forum on Facebook and Google+
  11. March 21st 2009, 06:42 AM #11
    stapel
    stapel is offline
    MHF Contributor
    Joined
    Mar 2007
    Posts
    1,240

    Talking

    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.
    Follow Math Help Forum on Facebook and Google+
  12. March 21st 2009, 11:23 AM #12
    HallsofIvy
    HallsofIvy is offline
    MHF Contributor
    Joined
    Apr 2005
    Posts
    13,727
    Thanks
    550
    I wouldn't consider that "skipping" any steps at all.
    Follow Math Help Forum on Facebook and Google+
« best strategy for this 3 variable equation? | Setting polynomial to zero »

Similar Math Help Forum Discussions

  1. Need help solving this equation
    Posted in the Trigonometry Forum
    Replies: 0
    Last Post: May 28th 2009, 05:08 AM
  2. I need help solving a equation
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: July 12th 2008, 01:34 PM
  3. solving diff equation w/ bournoulli's equation
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 14th 2008, 06:06 PM
  4. Solving a Log Equation
    Posted in the Algebra Forum
    Replies: 8
    Last Post: September 30th 2007, 09:22 AM
  5. Need help solving equation
    Posted in the Algebra Forum
    Replies: 2
    Last Post: September 28th 2007, 10:04 PM

Search Tags


/mathhelpforum @mathhelpforum