Results 1 to 11 of 11

Math Help - Solve for the given variable:

  1. #1
    Newbie
    Joined
    Sep 2009
    Posts
    10

    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:


    <br />
\frac{3x+4}{x-2} = \frac{5}{6}<br />

    And this is the second one:

    <br />
8t-\frac{6t-1}{2} = 5<br />

    Sorry, thanks for the help in advance!
    Last edited by Somatic; September 28th 2009 at 08:42 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,234
    Thanks
    27
    3x+\frac{4}{x}-2 = \frac{5}{6}

    Multiply both sides through by x

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

    That removed the fraction part in the second term.

    Now multiply through by 6

    18x^2+24-12x = 5x

    18x^2+24-17x = 0

    now you have a quadratic, can you solve this?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2009
    Posts
    10
    Thanks for the quick reply but I think you misunderstood my problem. I apologize, I wrote it incorrectly.
    This is how it should look:

    <br />
\frac{3x+4}{x-2} = \frac{5}{6}<br />

    And this is the second one:

    <br />
8t-\frac{6t-1}{2} = 5<br />

    I apologize, thank you for the help though.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,823
    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 (a) for a minute, the I have

    8t-6t-(a)=5

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

    2t-(a)=5

    Now let's add (a) to both sides

    2t=5+(a)

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

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

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

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


    Now, simplify the numerator

    \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

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

    Therefore

    t=\frac{11}{4}

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

    Bye!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Sep 2009
    Posts
    10
    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,234
    Thanks
    27
     \frac{3x+4}{x-2} = \frac{5}{6}

     6(3x+4) = 5(x-2)



    Now expand and group like terms.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Sep 2009
    Posts
    10
    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:

    x = \frac{-34}{13}

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

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

    I canceled the 2 by multiplying then I multiplied the other numbers as well.
    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.
    10t+1 = 10

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


    Now I have another problem that really throws me off.

    This:
    \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!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,234
    Thanks
    27
    Quote Originally Posted by Somatic View Post

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

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

    I canceled the 2 by multiplying then I multiplied the other numbers as well.
    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.
    10t+1 = 10

    End result :
    t = \frac{9}{10}
    Sounds good to me
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,234
    Thanks
    27
     \frac{4}{x} - \frac{x}{8} = 0

    multiply both sides by x gives

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

    Can you solve this?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Sep 2009
    Posts
    10
    Quote Originally Posted by pickslides View Post
     \frac{4}{x} - \frac{x}{8} = 0

    multiply both sides by x gives

     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:

     4 - x^2 = 0

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

     -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.

     x = - \sqrt(4)
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,234
    Thanks
    27
    You had the right idea, but didn't execute it too well.

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

    both sides by 8 gives

     32 - x^2 = 0

    then

     32 = x^2

     x^2 = 32

     x = \pm\sqrt{32}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Solve for the indicated variable:
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: August 8th 2011, 01:59 AM
  2. Solve for the specified variable?
    Posted in the Algebra Forum
    Replies: 2
    Last Post: March 4th 2011, 10:56 AM
  3. Solve for variable using d=rt ?
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 2nd 2009, 05:19 PM
  4. Solve for the variable...
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 30th 2009, 10:03 PM
  5. solve for specified variable
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 28th 2008, 09:24 PM

Search Tags


/mathhelpforum @mathhelpforum