Results 1 to 4 of 4

Math Help - Been a while, need a little help

  1. #1
    Newbie
    Joined
    Aug 2008
    Posts
    2

    Been a while, need a little help

    It has been a long time since math classes, however I need some algebra for a website I am working on. I am pretty much lost as how to get the number I need. So, here is an equation I am trying to solve:

    D = B x .80

    A = 120 + B + (D x 2) + ((D x.80) x 2)

    I am absolutely horrible with explanations so I made a diagram explaining what I am trying to accomplish.



    I need a solution in the form of B = A ......... I will know what A is at runtime, so if I can find a function I can create B and then create the rest of the objects. The problem is that A will change based on the users screen size, so I need a function like this to come up with B based on that.

    Thanks for any help, i'm pretty lost on this one.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Aug 2008
    From
    Chicago, IL
    Posts
    73
    D = B x .80

    A = 120 + B + (D x 2) + ((D x.80) x 2)

    solving for B...

    1st eliminate D
    A = 120 + B + ((B x .80) x 2) + ((B x .80) x 2)
    2nd eliminate inner parenthesis
    A = 120 + B + (B x 1.6) + (B x 1.6)
    3rd combine
    A = 120 + B + 2(B x 1.6)
    4th expand
    A = 120 + B + 2B x 3.2
    5th put together like terms
    A = 3B + 123.2
    and then basic algebra
    3B = A - 123.2
    B = (A - 123.2) / 3
    and thats it!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2008
    Posts
    2
    I appreciate the quick reply, and I think that it's really close. The only thing I noticed was in the first step where you plug B x .80 into for D.

    The second set of boxes is reduced 20% from the first set, which is reduced 20% from B.

    So I think with the D's eliminated it would look like

    A = 120 + B + ((B x .80) x 2) + (((B x .80) x .80) x 2)

    I came up with this (with help from you on how to eliminate the extra parenthesis)

    A = 120 + B + ((B x .80) x 2) + (((B x .80) x .80) x 2)

    A = 120 + B + (2B x 1.6) + ((.8B x .64) x 2)

    A = 120 + B + (2B x 1.6) + (1.6B x 1.28)

    Not sure where to go from there, and not 100% sure thats right. How do I add the two functions in the parenthesis together so I can start solving for B?

    Thanks again!

    [edit 2]

    I tried again, and I continued working on the problem, although the numbers I got don't seem to logically be possible.

    From the starting point above:

    A = 120 + B + 3.2B + 2.048B

    A = 120 + 6.248B

    A - 120 / 6.248 = B

    The 6.248 seems really inflated, i'm not very sure it's correct.

    [edit 3]

    I ran some numbers through the function that I got, and it spits out to low of a value for B, i'm just not sure where I messed up.

    [edit 4]

    I fixed it!! I did the distribution wrong, I looked at how you did it again and I got the function working!

    I am curious though, for example ((B x .80) x 2) why do you not distribute the 2 to the B as well as the .80?
    Last edited by Sileo; August 8th 2008 at 10:30 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Aug 2008
    From
    Chicago, IL
    Posts
    73
    A = 120 + B + (2B x 1.6) + (1.6B x 1.28)

    think of it as
    A = 120 + B + (2 x B x 1.6) + (1.6 x B x 1.28)
    so..
    A = 120 + B + (2 x B x 1.6) + (1.6 x B x 1.28)
    A = 120 + B + (3.2 x B) + (2.05 x B)
    then you can add the similar terms together
    A = 120 + B(1 + 3.2 + 2.05)
    A = 120 + 6.25 x B

    then simple algebra from there like last time.
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum