1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Solving for x (this should be easy but somehow I keep messing it up!)

  1. Aug 11, 2012 #1
    I'm starting college this fall. I asked one of my professors if we had a textbook and he said that there's an online version and I should read a bit of it if I wanted to.
    On the first chapter they were talking about simple (I mean really simple algebra, this is a vector geometry class) but I got completely stumped on one of the problems.

    I got an answer but it seems to be different from the textbook answer. I tried putting in number valves for the variable in the 2 different solution but i got different solutions. I probably did this wrong s0 can someone help me solve this I don't want to be stumped on my first week of college with the engineers >.<

    It should be uploaded on this message as an attachment.

    What I Did:

    called "problem my way" attachment

    Supposed Answer: attachment called "answer"

    Attached Files:

  2. jcsd
  3. Aug 11, 2012 #2


    User Avatar
    Homework Helper

    You're answer is the same as the given solution. You just have to clean your expression up a bit.
    [tex]x = \frac{e}{\frac{c - a}{a + 1/b} - d} - c[/tex]
    Take the fraction in the denominator and multiply by [itex]\frac{b}{b}[/itex]
    [tex]x = \frac{e}{\frac{b(c - a)}{ba + 1} - d} - c[/tex]
    Perform the subtraction in the denominator by finding the LCD (which is ba + 1):
    [tex]x = \frac{e}{\frac{b(c - a)}{ba + 1} - \frac{d(ba + 1)}{(ba + 1)}} - c[/tex]
    [tex]x = \frac{e}{\frac{b(c - a) - d(ba + 1)}{ba + 1}} - c[/tex]
    Multiply the entire fraction by [itex]\frac{ba + 1}{ba + 1}[/itex]
    [tex]x = \frac{e(ba + 1)}{b(c - a) - d(ba + 1)} - c[/tex]

    Note to mods: I don't think I'm giving too much away here. This is not in the HW subforum and the OP did show a lot of the work.
  4. Aug 12, 2012 #3
    I find it useful to do some substitutions when solving these, you could clean up the expression given and make things a lot easier
  5. Aug 14, 2012 #4
    Oh, THANK YOU!!! I tried using substitution to check but yeah since it's so messy I probably messed up somewhere. Thank you so much!!! This problem was really starting to freak me out. Thank you, eumayang, for cleaning it up! I tried converting my answer to the book's answer but I was really confused on how to do it.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook