Rational expressions.

Discussion in 'Undergraduate Math' started by Qast, Jan 20, 2007.

  1. Qast

    Qast Guest

    Hello,

    I am currently in a college "elementary" algebra course. I have never really done algebra before, so this is quite a shock for

    me. I am having a serious problem with "rational expressions" (mainly with variables).

    I sort of understand what it is that you do when you have variables in the denominator, but I have no idea why and I just

    simply do not comprehend it.

    For example, when you have:
    1 1
    ----- + -----
    x y

    I know the answer becomes:
    y + x
    --------
    xy

    But when you have, for example:
    2x + 4y
    ----- -----
    3xy cd

    I sort of understand what you do here as well, but not really. I know it will become 2xcd and I *think* 4y will become 4y^2x,

    but, again, I have no idea why.

    I also do not understand one bit how you end up reducing the problem. Do you reduce prior to combining everything--through

    prime factorization or what?

    Really what I need is a simple, easy-to-understand explanation of how you deal with these problems. I am becoming quite

    nervous, because I know manipulation of these is likely to become important as I continue my education (polynomials and

    such).

    Thanks for any help you can provide. I know this seems rather simple to many of you, but I'm struggling. My teacher just glossed right over this part and didn't explain it at all.
     
    Qast, Jan 20, 2007
    #1
    1. Advertisements

  2. Qast

    Qast Guest

    I'm sorry about the messed up formatting. Notepad is not very nice sometimes. Here it is again:

    Hello,

    I am currently in a college "elementary" algebra course. I have never really done algebra before, so this is quite a shock for me. I am having a serious problem with "rational expressions" (mainly with variables).

    I sort of understand what it is that you do when you have variables in the denominator, but I have no idea why and I just simply do not comprehend it.

    For example, when you have:
    1 1
    ----- + -----
    x y

    I know the answer becomes:
    y + x
    --------
    xy

    But when you have, for example:
    2x + 4y
    ----- -----
    3xy cd

    I sort of understand what you do here as well, but not really. I know it will become 2xcd and I *think* 4y will become 4y^2x, but, again, I have no idea why.

    I also do not understand one bit how you end up reducing the problem. Do you reduce prior to combining everything--through prime factorization or what?

    Really what I need is a simple, easy-to-understand explanation of how you deal with these problems. I am becoming quite nervous, because I know manipulation of these is likely to become important as I continue my education (polynomials and such).

    Thanks!
     
    Qast, Jan 20, 2007
    #2
    1. Advertisements

  3. Qast

    Paul Sperry Guest

    Two things you need to know:

    (1) a/b = ac/bc. In other words, it is OK to multiply (or divide)
    _BOTH_ the top and bottom by the same thing;

    (2) a/b + c/b = (a + c)/b; as long as the two denominators are the
    _SAME_ you can just add numerators.

    So, your strategy is to use (1) to get the denominators the same so
    that you can use (2).

    For example 1/x + 1/y = (yz)/(xyz) + (xz)/(xyz) --- using (1).

    Now you can use (2): 1/x + 1/y = (yz)/(xyz) + (xz)/(xyz) =
    (yz + xz)/(xyz).

    You can simplify if you want by dividing top and bottom by z to get
    1/x + 1/y = (yz + xz)/(xyz) = (z(x + y))/(xyz) = (x + y)/(xy).

    If you are wondering where the "z" came from (as you should be), I just
    tossed it in for the heck of it.

    (2x)/(3xy) + (4y)/(cd) = (2xcd)/(3xycd) + (4y3xy)/(cd3xy) =
    ((2xcd) + (12xy^2))/(3cdxy) = (x(2cd +12y^2))/(x3cdy) =
    (2cd + 12y^2)/(3cdy).

    Most of us would have started with (2x)/(3xy) = 2/(3y) but that is not
    required.

    Another example:

    1/(xy) + 1/(xz) = (xz)/(xyxz) + (xy)/(xzxy) = (xz + xy)/(xyxz) =
    (x(z + y))/(x(xyz)) = (z + y)/(xyz).

    Some may _insist_ on:
    1/(xy) + 1/(xz) = z/(xyz) + y/(xyz) = (z + y)/(xyz).

    Finally, do NOT write (xy + z)/(xw) = (y + z)/w. That guarantees big
    red marks on you homework.
     
    Paul Sperry, Jan 20, 2007
    #3
  4. Qast

    Temp Guest

    It is much the same as arithmetic with fractions. When two or more
    fractions have a common denominator, the numerators can be combined
    over the common denominator.

    1/2 + 1/3 = 3/6 + 2/6 = (3 + 2)/6 = 5/6

    In your first example, the lowest common denominator (LCD) = xy.
    Then, since xy/x = y and xy/y = x, multiply the first term by y/y and
    the second term by x/x.

    (y/y)*(1/x) + (x/x)*(1/y) = y/xy + x/xy = (y + x) / (xy)


    In your second example, you could take the LCD to be 3xycd. But, in
    this case, we can simplify things a bit by noticeing the common factor
    'x' in the first term, which we can cancel, leaving

    2 4y
    -- + --
    3y cd

    Now the LCD is 3ycd. Since 3ycd / 3y = cd and 3ycd / cd = 3y,

    (cd / cd)*(2 / 3y) + (3y / 3y)*(4y / cd)

    = 2cd / 3ycd + 12 y^2 / 3ycd = (2cd + 12y^2) / (3ycd)


    Finally, notice that in the special case where you have *two* terms,
    you can always simply 'cross multiply'.

    a b
    - + - = (ad + bc) / (cd)
    c d

    Imagine an "X" connecting a and d, and b and c, hence the name "cross
    multiply".


    Let's do your second example again, and this time we won't cancel the
    'x' first, but we will simply cross multiply.

    2x 4y
    --- + --
    3xy cd

    2xcd + 12 x y^2
    = ---------------
    3xycd

    Now, notice that there is a common factor 'x' everywhere, which we can
    cancel, leaving

    2cd + 12 y^2
    = ------------
    3ycd
     
    Temp, Jan 20, 2007
    #4
  5. Qast

    Qast Guest

    Thanks for your responses!
     
    Qast, Jan 21, 2007
    #5
  6. Yes, but do you understand WHY?????

    ALL Arithmetic with fractions depends on one thing: THE DISTRIBUTIVE
    LAW.
    If you understand nothing else about fractions, you can also go back
    to the
    distributive law to derive your answer. Virtually all algebraic
    manipulations that
    you will encounter in your course stem from this law. LEARN IT WELL.
    UNDERSTAND IT.

    We have A(B+C) = AB + AC

    We can't add fractions directly. What we can do is add integers, then
    pull out the
    common factor from the integers we are adding.

    You first need to find A. To do this, we find the LEAST COMMON
    MULTIPLE of x and y

    [or the least common multiple of all terms in all the denominators of
    the fractions we are
    adding]
    which is x*y. Now express 1/x and 1/y as AB and AC
    respectively giving
    A = 1/xy, B = y and C = x.

    Now factor out A from AB + AC giving 1/xy( x + y)

    You are trying to do too much by applying "formulae" and not paying
    attention to basic
    principles.

    The LCM of 3xy and cd is 3xycd. Take A = 3xycd. Now B
    = 2xcd and C = 4y * 3xy

    Whence the sum is 1/(3xycd) ( 2xcd + 12xy^2)


    This is no different than adding (say) 12/143 + 16/91. The only
    difference is that
    you are not comfortable with the use of VARIABLES to represent numbers.

    You must get comfortable with the notion of variables (and how to
    substitute one variable for another)
    to have any hope of learning algebra.
     
    Pubkeybreaker, Jan 22, 2007
    #6
    1. Advertisements

Ask a Question

Want to reply to this thread or ask your own question?

You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.