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Thread: Solve for the given variable:

  1. #1
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    Solve for the given variable:

    I'm having trouble with these problems. Every time I see a fraction I get lost in these kind of equations.

    Here is my first equation:

    3x+4/x-2 = 5/6

    and here is the second one:

    8t - 6t-1/2 = 5

    Thanks in advance.

    EDIT:
    I apologize, I wrote my problems wrong!

    This how they should be:


    $\displaystyle
    \frac{3x+4}{x-2} = \frac{5}{6}
    $

    And this is the second one:

    $\displaystyle
    8t-\frac{6t-1}{2} = 5
    $

    Sorry, thanks for the help in advance!
    Last edited by Somatic; Sep 28th 2009 at 08:42 PM.
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  2. #2
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    $\displaystyle 3x+\frac{4}{x}-2 = \frac{5}{6}$

    Multiply both sides through by x

    $\displaystyle 3x^2+4-2x = \frac{5x}{6}$

    That removed the fraction part in the second term.

    Now multiply through by 6

    $\displaystyle 18x^2+24-12x = 5x$

    $\displaystyle 18x^2+24-17x = 0$

    now you have a quadratic, can you solve this?
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  3. #3
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    Thanks for the quick reply but I think you misunderstood my problem. I apologize, I wrote it incorrectly.
    This is how it should look:

    $\displaystyle
    \frac{3x+4}{x-2} = \frac{5}{6}
    $

    And this is the second one:

    $\displaystyle
    8t-\frac{6t-1}{2} = 5
    $

    I apologize, thank you for the help though.
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Somatic View Post
    I'm having trouble with these problems. Every time I see a fraction I get lost in these kind of equations.

    Here is my first equation:

    3x+4/x-2 = 5/6

    and here is the second one:

    8t - 6t-1/2 = 5

    Thanks in advance.
    Fractions are our friends!

    Just think of them as numbers. Look, if you don't like the fractions, replace it with a letter until the rest of the problem is done. Then, when you solve for x, put the fraction where it's supposed to go. It's like a crutch until you get it.

    Check out the second problem.

    if I pretend that 1/2 is $\displaystyle (a)$ for a minute, the I have

    $\displaystyle 8t-6t-(a)=5$

    Now, we want to get t by itself, so notice that $\displaystyle 8t$ and $\displaystyle 6t$ are like terms, so we can combine them like this

    $\displaystyle 2t-(a)=5$

    Now let's add $\displaystyle (a)$ to both sides

    $\displaystyle 2t=5+(a)$

    Now we divide both sides by two, and we have t alone at last

    $\displaystyle t=\frac{5+(a)}{2}$

    Now would be a good time to put the fraction back where $\displaystyle (a)$ is

    $\displaystyle t=\frac{5+1/2}{2}$


    Now, simplify the numerator

    $\displaystyle \frac{5+1/2}{2}=\frac{\frac{10}{2}+\frac{1}{2}}{2}=\frac{\fr ac{11}{2}}{2}$

    Now, I know this look crazy, but if you think about it

    $\displaystyle \frac{\frac{11}{2}}{2}=\frac{11}{2}\div\frac{2}{1} =\frac{11}{2}\cdot\frac{1}{2}=\frac{11}{4}$

    Therefore

    $\displaystyle t=\frac{11}{4}$

    This may confuse the heck out of you, but its another way to do things.

    Bye!
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  5. #5
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    Sorry guy!
    I edited my first post, they have the correct equation. I wasn't sure on how to use the math tags. I'm sorry if I confused you all.
    Thanks for the help though, I appreciate it!

    Can someone give me a start on my problems please, thank you for the above posters for the help either way.
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  6. #6
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    $\displaystyle \frac{3x+4}{x-2} = \frac{5}{6}$

    $\displaystyle 6(3x+4) = 5(x-2)$



    Now expand and group like terms.
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  7. #7
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    Wow, thank you pickslides. I tend to over think things when I don't know how to begin these types of equations. I got my solution to the first problem which is:

    $\displaystyle x = \frac{-34}{13}$

    I am trying to solve the next one but I think I'm doing a step wrong.

    $\displaystyle 8t - \frac{6t-1}{2} = 5$

    I canceled the 2 by multiplying then I multiplied the other numbers as well.
    $\displaystyle 16t-6t+1 = 10$

    The 1 changes to a positive correct?

    Am I on the right track?

    EDIT:

    I solved it.
    I continued and did this.
    $\displaystyle 10t+1 = 10$

    End result :
    $\displaystyle t = \frac{9}{10}$


    Now I have another problem that really throws me off.

    This:
    $\displaystyle \frac{4}{x} - \frac{x}{8} = 0$

    Just give me a step please I want to try to solve but I don't know where to start.
    Thanks for the support!
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  8. #8
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    Quote Originally Posted by Somatic View Post

    I am trying to solve the next one but I think I'm doing a step wrong.

    $\displaystyle 8t - \frac{6t-1}{2} = 5$

    I canceled the 2 by multiplying then I multiplied the other numbers as well.
    $\displaystyle 16t-6t+1 = 10$

    The 1 changes to a positive correct?

    Am I on the right track?

    EDIT:

    I solved it.
    I continued and did this.
    $\displaystyle 10t+1 = 10$

    End result :
    $\displaystyle t = \frac{9}{10}$
    Sounds good to me
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  9. #9
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    $\displaystyle \frac{4}{x} - \frac{x}{8} = 0$

    multiply both sides by x gives

    $\displaystyle 4 - \frac{x^2}{8} = 0$

    Can you solve this?
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  10. #10
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    Quote Originally Posted by pickslides View Post
    $\displaystyle \frac{4}{x} - \frac{x}{8} = 0$

    multiply both sides by x gives

    $\displaystyle 4 - \frac{x^2}{8} = 0$

    Can you solve this?
    Okay, I'm not sure if I'm doing this correctly.

    I then multiplied both sides by 8 giving me this:

    $\displaystyle 4 - x^2 = 0 $

    After this continued with adding 4 to both sides got this:

    $\displaystyle -x^2 = 4 $

    Is this my answer? I'm pretty sure I'm off big time. Please elaborate on what I did wrong, if I am incorrect.

    $\displaystyle x = - \sqrt(4) $
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  11. #11
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    You had the right idea, but didn't execute it too well.

    $\displaystyle 4 - \frac{x^2}{8} = 0$

    both sides by 8 gives

    $\displaystyle 32 - x^2 = 0$

    then

    $\displaystyle 32 = x^2 $

    $\displaystyle x^2 = 32$

    $\displaystyle x = \pm\sqrt{32}$
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