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  1. #1
    Guest

    Help with Algebra??

    This question is off topic. I have to take a Placement Test for college to see what classes that should put me in. So i got a library book on forgotten algebra and i need some help. I will do the best i can to write out these problems because i do not have a fancy algebra, trig and calculus and so on keyboard. Here are the problems as follow. Before i start. The are a few things. I need to know how you got the answer. Has anyone did math long distance over the internet before??

    5-2x
    ------- = -2
    x-1

    2
    -- + 3(x-1)
    x -------- =1
    5x

    4
    ---- - 1
    x-2 ---- =5
    x ------
    x-2


    "There are 3 separate problems. I think you have to cross mutiply. I had to use the hyphen mark so it's all mess up and side ways. Please reply back. Thanks for any help."




  2. #2
    Mentor
    Join Date
    Jun 2001
    Posts
    1,672

    Re:Help with Algebra??

    These seem to be very simple problems, however it is very hard to figure out how the problem is layed out, especially on the last two. I know it is very hard to do using simple characters. Can you write them out lined up a little better?

    The first one should go like this:

    5-2x
    ------- = -2
    x-1

    Multiply both sides by (x-1):
    5-2x = -2(x-1)
    Let's break up the paranthesis on the right hand side:
    5-2x = -2x - (-2)
    5-2x = -2x + 2
    Let's add 2x to both sides:
    5 = 2

    It has no solution ...

  3. #3
    Guest

    Re:Help with Algebra??

    Do you have a scan of these problems instead - I highly doubt ACSII formated by the forum is a good solution.

    anyways Aleabra is very simple, the approach cloverm showed you will solve any algebra problem on that level.

  4. #4

    Re:Help with Algebra??

    [quote author=cloverm link=board=14;threadid=7355;start=#msg67754 date=1057807119]
    These seem to be very simple problems, however it is very hard to figure out how the problem is layed out, especially on the last two. I know it is very hard to do using simple characters. Can you write them out lined up a little better?

    The first one should go like this:

    5-2x
    ------- = -2
    x-1

    Multiply both sides by (x-1):
    5-2x = -2(x-1)
    Let's break up the paranthesis on the right hand side:
    5-2x = -2x - (-2)
    5-2x = -2x + 2
    Let's add 2x to both sides:
    5 = 2

    It has no solution ...

    [/quote]

    Ok I may have done this wrong, but I may have to disagree with you cloverm. I agree up to this part:
    5-2x = 2x-2

    But I would do this from here on out:
    5 - 2x +2 = 2x - 2 +2
    7-2x=2x
    7-2x+2x=2x+2x
    7=4x
    7/4=x

    Like I said this has a very high likely hood of being wrong, its been a LONG TIME.

    <edit>
    Ok I was very wrong! :-[ I checked my solution, it doesn't work out! You are the master cloverm!! ;D
    </edit>

  5. #5
    Mentor
    Join Date
    Jun 2001
    Posts
    1,672

    Re:Help with Algebra??

    [quote author=Izan Seth link=board=14;threadid=7355;start=#msg67766 date=1057846516]

    Ok I may have done this wrong, but I may have to disagree with you cloverm. I agree up to this part:
    5-2x = 2x-2

    [/quote]
    No, no, you left out the - sign on the right hand side:

    5-2x = -2(x-1)

  6. #6

    Re:Help with Algebra??

    Ah yes that would change things quite a bit. Like I said I'm just dumb! :-[ :-[ :-[ My appologies!

  7. #7
    Guest

    Re:Help with Algebra??

    Thanks guys. Here are the other 2 problems. I will do my best to line them up. I have not had algebra in about 2 years. The last algebra i had was Algebra 1.

    2 3(x-1)
    --- + -------- =1
    x 5x


    4 1
    ----- - ----- = 5
    x-2 x ------
    x-2


    "How is that?? Please reply back. Thanks."

  8. #8
    Guest

    Re:Help with Algebra??

    If I'm correct:

    1)

    Contraction of division:
    10x/5x^2 + (3(x-1)x)/5x^2 = 1
    (10x + 3(x-1)x) / 5x^2 = 1

    Isolate x
    (3x^2 + 7x) / 5x^2 =1
    3x^2 + 7x = 5x^2
    3x + 7 = 5x
    -2x = -7

    x = (7/2)

    control:
    2/(7/2) + (3((7/2) - 1)) / 5(7/2)) = 1

    ~0.57 + ~0.43 = 1

    This seems correct!

    2)
    Same as before:

    Contraction:
    4/(x-2) - 1/x = 5/(x-2)

    4/(x-2) - 5/(x-2) = 1/x

    -1 / (x-2) = 1/x

    -1 = (x-2) / x
    -1 = 1 - 2/x
    -2 = -2/x
    -2x = -2
    x = 1

    control:

    4 / ( -1) - (1 / 1) = 5 / ( -1)
    -4 -1 = -5
    -4 = - 4

    this also seems correct

    -edit-

    oh and damn, you made me do math during my vaction.. I had to think... that hurt.



  9. #9
    Guest

    Re:Help with Algebra??

    Lovechild. How the hell did you do that?? Now my head hurts because you have me thinking. Is there an easier way around these. I have this feeling when i take the test they are going to place me in arithmetic because i had not take any algebra in about 2 years. I'm seriously going brain dead. Please show me an easier way. Please reply back. Thanks Lovechild.

  10. #10
    Guest

    Re:Help with Algebra??

    Okay basically the first step is always to contract the multiple divisions into one common division (by getting a common divisor)

    2/x + 1/y would then be (2*y)/(x*y) + (1*x)/(y*x)
    and contracted it would be (2y + x) / xy

    This way you can proced pretty much like you would always do, this operation is legal because you infact only extend the division by 1 (x/x is always 1, as is y/y)

    Always extend the one division by the others divisor and vice virsa to get a common divisor . and thus be able to get only one division to work with.

    now is 2/x + 1/y = 2
    then

    (2y + x) / xy = 2
    2y +x = 2xy

    and so and and so forth (disgard the fact that is has infinte outcomes due to the fact that you have multiple variables - it's just to show how it's done).

    the rest is simply to isolate the variable (if you have more variables this calls for some more magic but that's no harder you basically need as many different equations as you have variables to solve that).

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