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Math Help - Solving equations

  1. #1
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    Solving equations

    ...(2) = ...(2x) + 3
    (x + 1) (2x - 3)

    The dots don't mean anything I am trying to make it look OK?

    I am asked to solve the above equation!

    I think my first step is to determine the denominators, so I think;

    common denominators are; 3(x+1)(2x-3)

    when I expand these denominators I get;

    3(x+1)(2x-3) = 2x^2 - 3x + 2x = 3(2x^2 - 3x +2x) = 6x^2 - 9x + 6x = 6x^2 - 3x

    So my common denominators are 3 and 6

    am I right so far?

    Thanks

    David
    Last edited by David Green; November 17th 2011 at 08:09 AM. Reason: maths layout does not look good
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  2. #2
    Super Member Quacky's Avatar
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    Re: Solving equations

    Is the problem statement:

    \frac{2}{x+1}=\frac{2x+3}{2x-3}?
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  3. #3
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    Re: Solving equations

    Quote Originally Posted by Quacky View Post
    Is the problem statement:

    \frac{2}{x+1}=\frac{2x+3}{2x-3}?
    nearly, I wished I could present the work as you do then I would not make things difficult for people?

    ........2x..... + 3
    ......2x - 3
    Sorry I hope this is clearer

    David
    Last edited by David Green; November 17th 2011 at 08:13 AM. Reason: just a struggle presenting it right
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  4. #4
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    Re: Solving equations

    this forum implements a math-formatting system called "latex". to write a fraction in latex, you type:

    [tex]\frac{expression one}{expression two}[/tex]

    so [tex]\frac{2x}{x^2+1}[/tex]

    produces \frac{2x}{x^2 + 1}.

    read this thread, it has more (note: we don't use the "math" tags anymore, we use "tex" tags).
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  5. #5
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    Re: Solving equations

    Quote Originally Posted by Deveno View Post
    this forum implements a math-formatting system called "latex". to write a fraction in latex, you type:

    \frac{expression one}{expression two}

    so \frac{2x}{x^2+1}

    produces \frac{2x}{x^2 + 1}.

    read this thread, it has more (note: we don't use the "math" tags anymore, we use "tex" tags).
    I am just going to try it copying your example.

    {tex}\frac{2x}{x^22+1}{tex}

    OK so that example for me didn't work, please advise?
    Last edited by David Green; November 17th 2011 at 08:44 AM. Reason: Not working
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  6. #6
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    Re: Solving equations

    use square brackets [], not curly ones {} around the tex tags. remember you start with [ tex] and end with [/tex]. the curly brackets seperate different things for the \frac function inside the tex tags.
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  7. #7
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    Re: Solving equations

    Quote Originally Posted by Deveno View Post
    use square brackets [], not curly ones {} around the tex tags. remember you start with [ tex] and end with [/tex]. the curly brackets seperate different things for the \frac function inside the tex tags.
    OK here I go?

    [tex]\frac[2x][x^2][2+1][tex]

    There must be something else involved?

    I still can't get it
    Last edited by David Green; November 17th 2011 at 09:07 AM. Reason: Not working
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  8. #8
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    Re: Solving equations

    [Tex]\frac{2x}{x^2+1}[\tex] But use / instead of \

    Anyway, so we have:

    \frac{2x}{x+1}=\frac{2x}{2x-3}+3 I think.

    There are a few ways that you could go about this. I'd start by collecting everything on one side of the equation:

    0=\frac{2x}{2x-3}+3-\frac{2x}{x+1}

    Now, you could, if you wanted to, find a common denominator. The lowest common denominator isn't actually affected by the 3 though because we can manipulate this to match anything we want. Just look at the algebraic fractions. (x+1)(2x-3) is the lowest common denominator for both of them. So I multiply both the numerator and denominator by the missing components, as it were, one at a time.


    0=\frac{2x{\color{red}(x+1)}}{(2x-3){\color{red}(x+1)}}+\frac{3{\color{red}(x+1)(2x-3)}}{{\color{red}(x+1)(2x-3)}}-\frac{2x{\color{red}(2x-3)}}{(x+1){\color{red}(2x-3)}}

    Hopefully you should be able to see that the parts in red cancel to give us exactly what we had in the stage above. I've multiplied through both the numerator and denominator by the same value, so I haven't changed anything - the denominators are now exactly the same. Can you combine and expand them? You need to rewrite everything as one fraction. You don't need to change the denominators at all.
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  9. #9
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    Re: Solving equations

    Quote Originally Posted by Quacky View Post
    [Tex]\frac{2x}{x^2+1}[\tex] But use / instead of \

    Anyway, so we have:

    \frac{2x}{x+1}=\frac{2x}{2x-3}+3 I think.

    There are a few ways that you could go about this. I'd start by collecting everything on one side of the equation:

    0=\frac{2x}{2x-3}+3-\frac{2x}{x+1}

    Now, you could, if you wanted to, find a common denominator. The lowest common denominator isn't actually affected by the 3 though because we can manipulate this to match anything we want. Just look at the algebraic fractions. (x+1)(2x-3) is the lowest common denominator for both of them. So I multiply both the numerator and denominator by the missing components, as it were, one at a time.


    0=\frac{2x{\color{red}(x+1)}}{(2x-3){\color{red}(x+1)}}+\frac{3{\color{red}(x+1)(2x-3)}}{{\color{red}(x+1)(2x-3)}}-\frac{2x{\color{red}(2x-3)}}{(x+1){\color{red}(2x-3)}}

    Hopefully you should be able to see that the parts in red cancel to give us exactly what we had in the stage above. I've multiplied through both the numerator and denominator by the same value, so I haven't changed anything - the denominators are now exactly the same. Can you combine and expand them? You need to rewrite everything as one fraction. You don't need to change the denominators at all.
    I have copied the info to do as you ask, but just one question regarding the RED bits which you advise cancel, should these now be cancelled or combined and expanded?

    Thanks

    David
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  10. #10
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    Re: Solving equations

    combined and expanded (and by this i mean: add the numerators). your main concern is finding out what the numerator over a common denominator will be, because THAT numerator has to equal 0.
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  11. #11
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    Re: Solving equations

    Quote Originally Posted by Quacky View Post
    [Tex]\frac{2x}{x^2+1}[\tex] But use / instead of \

    Anyway, so we have:

    \frac{2x}{x+1}=\frac{2x}{2x-3}+3 I think.

    There are a few ways that you could go about this. I'd start by collecting everything on one side of the equation:

    0=\frac{2x}{2x-3}+3-\frac{2x}{x+1}

    Now, you could, if you wanted to, find a common denominator. The lowest common denominator isn't actually affected by the 3 though because we can manipulate this to match anything we want. Just look at the algebraic fractions. (x+1)(2x-3) is the lowest common denominator for both of them. So I multiply both the numerator and denominator by the missing components, as it were, one at a time.


    0=\frac{2x{\color{red}(x+1)}}{(2x-3){\color{red}(x+1)}}+\frac{3{\color{red}(x+1)(2x-3)}}{{\color{red}(x+1)(2x-3)}}-\frac{2x{\color{red}(2x-3)}}{(x+1){\color{red}(2x-3)}}

    Hopefully you should be able to see that the parts in red cancel to give us exactly what we had in the stage above. I've multiplied through both the numerator and denominator by the same value, so I haven't changed anything - the denominators are now exactly the same. Can you combine and expand them? You need to rewrite everything as one fraction. You don't need to change the denominators at all.
    I don't think I will be anywhere near right to be honest, and if I put my full working out in now I am sure it would be very confusing?

    However I have ended up with;

    8^2 - 7 - 2x / 2x^2 - x - 7

    ????????
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  12. #12
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    Re: Solving equations

    Just having another go at latex?

    /frac{2x}{x^22+1}

    [tex\frac{2x}{x^2}{2+1}[/tex]
    Last edited by David Green; November 17th 2011 at 10:05 AM. Reason: still learning
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  13. #13
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    Re: Solving equations

    In the future, would you mind posting some working out stages rather than jumping immediately several stages ahead? It would help us to follow your logic and check your solution.

    Firstly, you have no reason to do anything with the denominator. Once you've found the common denominator, just leave it as it is. Expanding it is just wasting your energy (and potentially exam time!).

    When we combine everything, we get:

    \frac{2x(x+1)+3(x+1)(2x-3)-2x(2x-3)}{(2x-3)(x+1)}

    This is certainly a monster, so don't try to rush it. Take it term by term. I'd suggest starting by rewriting the 3(x+1) towards the middle as (3x+3)

    Then, start chronologically from the beginning and expand, using FOIL if necessary.
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  14. #14
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    Re: Solving equations

    Quote Originally Posted by David Green View Post
    I don't think I will be anywhere near right to be honest, and if I put my full working out in now I am sure it would be very confusing?

    However I have ended up with;

    8^2 - 7 - 2x / 2x^2 - x - 7

    ????????
    The numerator is 2x(x+1) + 3(2x-3)(x+1) - 2x(2x-3). Expand and collect like terms (and be aware of the rule regarding subtracting negatives in the last term)

    The denominator is best left as (2x-3)(x+1) since we'll multiply both sides by this to clear the fraction and since anything multiplied by 0 is 0 we need only look at the numerator
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  15. #15
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    Re: Solving equations

    Quote Originally Posted by David Green View Post
    Just having another go at latex?

    /frac{2x}{x^22+1}
    This is fine except that your slash is the wrong direction in the "/frac" - it should be "\frac" and I think it should be x^2 instead of x^22
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