Results 1 to 7 of 7

Math Help - Solving for X (new to forums)

  1. #1
    Newbie
    Joined
    Jan 2010
    Posts
    15

    Solving for X (new to forums)

    I am reviewing math in the back of my physics book and I don’t have a college algebra book anymore. I tried goggling this for 2 hours or so and could not find anything similar. It seems like it’s a certain type of problem giving me problems.

    Two problems I have the answer to both but cannot figure out how they came up with the answer.

    Solve for x

    ax – 5 = bx +2 Answer is: x = (7/(a-b))

    I cannot figure this out I moved the 5 over that’s easy

    ax = bx + 7

    I then tried to move the “bx”

    ax – bx = 7

    I then divided 7 by -b

    ax –x = (7/(-b))

    I then divided by a

    x – x = (7/(a-b))

    However would not x – x = 0

    I must not be getting some rule…. Even if when I moved the negative b over if I didn’t leave a neg it would be x +x which would = 2x
    Right? I am confused on this and the other one I had problems with Is similar.

    A = (1/(1+x)) Answer is x = ((1-a)/(a))

    I think these are related because this one also I could not solve this one either… I solve all the normal x stuff but it has to be a rule I am missing…




    I think I figured out the second problem as listed below I left it in just in case I made a mistake.. I have taken College Algebra, Trig, Calculus I but that was back in 1997. I am now enrolled in Engineering Physics, and Calculus II which start next week so trying to review and catch up and review stuff on my own before I am really hazy on this stuff so forgive any seeming ignorance on my part. I may be posting a lot of questions on here in the coming weeks. Next week I will have access to school tutors as well and I should be getting in my Calculus I, and II MathTutorDvD I ordered from a website... hopefully they are good if so I will order Calc III and Physics CD from them as well. I use to be really good at math but not feeling so good at it right now

    a = (1/(1+x))

    took (1+x) and since I was dividing I did the opposite and multiplied by both sides

    a(1+x) = 1

    I think that then breaks down to

    1a + ax = 1

    I then did opposite moving 1a over

    ax = 1 – a (since “a” and “1a” are the same thing I just took labeled it as “a”)

    I then divided by a

    X = (1- a)/(a)

    I don’t know why I got confused on that one now it seems easy


    So its just that first one I am having a problem with ugggg I hate not being able to figure out something I will almost go till my nails are fingers are bleeding and I fall over in the chair from exhaustion lol. Learning to put things away to come back a different day is very hard for me lol
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member I-Think's Avatar
    Joined
    Apr 2009
    Posts
    288
    ax-5=bx+2

    ax=bx+2+5

    ax=bx+7

    ax-bx=7

    x(a-b)=7

    x=\frac{7}{a-b}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2010
    Posts
    11
    Quote Originally Posted by wilder7bc View Post
    I am reviewing math in the back of my physics book and I don’t have a college algebra book anymore. I tried goggling this for 2 hours or so and could not find anything similar. It seems like it’s a certain type of problem giving me problems.

    Two problems I have the answer to both but cannot figure out how they came up with the answer.

    Solve for x

    ax – 5 = bx +2 Answer is: x = (7/(a-b))

    I cannot figure this out I moved the 5 over that’s easy

    ax = bx + 7

    I then tried to move the “bx”

    ax – bx = 7

    I then divided 7 by -b

    ax –x = (7/(-b))

    I then divided by a

    x – x = (7/(a-b))

    However would not x – x = 0

    I must not be getting some rule…. Even if when I moved the negative b over if I didn’t leave a neg it would be x +x which would = 2x
    Right? I am confused on this and the other one I had problems with Is similar.

    A = (1/(1+x)) Answer is x = ((1-a)/(a))

    I think these are related because this one also I could not solve this one either… I solve all the normal x stuff but it has to be a rule I am missing…




    I think I figured out the second problem as listed below I left it in just in case I made a mistake.. I have taken College Algebra, Trig, Calculus I but that was back in 1997. I am now enrolled in Engineering Physics, and Calculus II which start next week so trying to review and catch up and review stuff on my own before I am really hazy on this stuff so forgive any seeming ignorance on my part. I may be posting a lot of questions on here in the coming weeks. Next week I will have access to school tutors as well and I should be getting in my Calculus I, and II MathTutorDvD I ordered from a website... hopefully they are good if so I will order Calc III and Physics CD from them as well. I use to be really good at math but not feeling so good at it right now

    a = (1/(1+x))

    took (1+x) and since I was dividing I did the opposite and multiplied by both sides

    a(1+x) = 1

    I think that then breaks down to

    1a + ax = 1

    I then did opposite moving 1a over

    ax = 1 – a (since “a” and “1a” are the same thing I just took labeled it as “a”)

    I then divided by a

    X = (1- a)/(a)

    I don’t know why I got confused on that one now it seems easy


    So its just that first one I am having a problem with ugggg I hate not being able to figure out something I will almost go till my nails are fingers are bleeding and I fall over in the chair from exhaustion lol. Learning to put things away to come back a different day is very hard for me lol
    Alright with the first one, what you did initially was correct, so I am going to start with

    ax – bx = 7

    What you need to do next is realize that both (ax) and (bx) have an (x) term, so you can factor

    ax - bx into x(a-b) I think that is the confusing part. once you do that
    x(a-b) = 7, simply divide both sides by (a-b)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jan 2010
    Posts
    15

    nice one question

    Wow... that was fast...

    I have a question though...

    I am not sure I understand this:



    Is there any documentation on now this works or what its called. I am trying to remember this but its not striking a bell.

    I understand looking at it that x(a - b) = 7 is the same as xa - xb = 7

    but I dont remember pulling an x out like that. (btw is that algebra I, II, or College Algebra? (my terminology is going to be yuck for awhile lol)

    Hey thanks a ton for the prompt answer I am so glad I found this forum!

    Sincerely,

    Brian
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jan 2010
    Posts
    15

    great!

    Ok I think S31J41 answered the second part of my question... I am so relieved I tried figuring that out for hours...

    So thats called factoring out and that definitely sounds familiar and I am starting to recall a tad bit about that. Also can now maybe do searches on it if I need to.

    Guys are awesome thanks!

    Sincerely,

    Brian
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Jul 2009
    From
    Melbourne
    Posts
    274
    Thanks
    4
    I'll just contribute a tad more (cause i can :P)

    For the second one you have

    a(1+x)=1

    You can simply divide the right hand side by a because  a(1+x) is  a * (1+x)

    1+x = \frac{1}{a}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member rowe's Avatar
    Joined
    Jul 2009
    Posts
    89
    Quote Originally Posted by wilder7bc View Post
    Wow... that was fast...

    I have a question though...

    I am not sure I understand this:



    Is there any documentation on now this works or what its called. I am trying to remember this but its not striking a bell.

    I understand looking at it that x(a - b) = 7 is the same as xa - xb = 7

    but I dont remember pulling an x out like that. (btw is that algebra I, II, or College Algebra? (my terminology is going to be yuck for awhile lol)

    Hey thanks a ton for the prompt answer I am so glad I found this forum!

    Sincerely,

    Brian
    This is called the Distributive Law, which, more generally is:

    x(a+b) = (xa+xb)

    Here are a couple of examples:

    2(x - 3) = (2x - 2\cdot 3) = (2x - 6)

    x(a^2 + x) = (xa^2 + x^2)

    (y+3a)(2 - 7a) = ((y+3a)2 - (y+3a)7a)
    = ((2y+6a) - (7ay + 21a^2))
    = 2y + 6a - 7ay - 21a^2

    Note that in the last example, (y+3a)2 = 2(y+3a) due to the Commutativity of multiplication.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. new to forums, and transforming formulas :D
    Posted in the Algebra Forum
    Replies: 5
    Last Post: November 19th 2008, 12:43 PM

Search Tags


/mathhelpforum @mathhelpforum