Page 1 of 2 12 LastLast
Results 1 to 15 of 22

Math Help - Rational Equations

  1. #1
    Newbie
    Joined
    Mar 2007
    Posts
    12

    Rational Equations

    Okay we're doing Rational Equations in my Algebra class. And I know that you have to find a common denominator but I'm not sure what to do after that. Here's an example equation:

    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Legend9591 View Post
    Okay we're doing Rational Equations in my Algebra class. And I know that you have to find a common denominator but I'm not sure what to do after that. Here's an example equation:

    When given expressions like these, you can take the multiple of the denominators as the common denominator. so the common denominator for 3/(x + 1) and 1/(x - 2) is (x + 1)(x - 2)

    Here's how to do the problem. There are two mainstream methods, the first is the one you are interested in.

    3/(x + 1) - 1/(x - 2) = 1/(x^2 - x - 2)
    3(x - 2) - 1(x + 1)................1
    -------------------- = ---------------
    .....(x + 1)(x - 2)..........(x^2 - x - 2)

    => ........3x - 6 - x - 1..................1
    ........-------------------- = ---------------
    .............(x + 1)(x - 2)..........(x^2 - x - 2)

    => ............2x - 7.......................1
    ........-------------------- = ---------------
    .............(x + 1)(x - 2)..........(x^2 - x - 2)

    notice that the denominators are the same, so the numerators must be the same.

    so 2x - 7 = 1
    => 2x = 1 + 7
    => x = 4


    The alternate method is this:
    notice that 1/(x^2 - x - 2) = 1/(x + 1)(x - 2) .............so the bottom can factor into two expressions, each of which are the denominator of a fraction on the other side of the equal sign. We can multiply by the denominator (x + 1)(x - 2)

    so we get:

    (x + 1)(x - 2)[3/(x + 1) - 1/(x - 2)] = (x + 1)(x - 2)[1/(x^2 - x - 2)]

    => 3(x - 2) - 1(x + 1) = 1
    => 3x - 6 - x -1 = 1
    => 2x - 7 = 1 ........................notice this is the same equation we ended up solving in the previous method
    => x = 4
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2007
    Posts
    12
    Ooh! Okay that makes so much more sense... Thank you!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Legend9591 View Post
    Ooh! Okay that makes so much more sense... Thank you!
    I edited what i wrote, so make sure to view it again to see if you saw the updated version. if there is anything in either method that you dont get, say so
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2007
    Posts
    12
    Which method would you suggest I use...? Which one is easier?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Legend9591 View Post
    Which method would you suggest I use...? Which one is easier?
    In terms of easiness, i'd say both are equal, however, I would suggest the second one, it is shorter and quicker to do, and hence a better method if you are taking a test. But if a question (or your professor) specifically mentions "common denominator", then you must do the first method.

    I dont suspect either method would be difficult for you to learn however. if you get both of them, then whatever you are required to do you can do it. Do you understand both methods?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Mar 2007
    Posts
    12
    The questions say: Factor first, then solve the equation.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Legend9591 View Post
    The questions say: Factor first, then solve the equation.
    In that case, either method can work, i'd use the second one. I suspect the "factor first" is refering to the denominator on the right hand side, they want you to factor x^2 - x - 2 as (x + 1)(x - 2) before doing the problem
    Follow Math Help Forum on Facebook and Google+

  9. #9
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Legend9591 View Post
    The questions say: Factor first, then solve the equation.

    I think you should learn both methods just to be safe. then you just know to always use the second one UNLESS the question asks you to find a "common denominator" or "combine the fractions"
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Mar 2007
    Posts
    12
    I'm confused by the second method of doing the equation.

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

    is [3/(x+1) - 1/(x-2)] right?

    But how did you 3(x-2) - 1(x+1) when the equation would be:

    Would the equation be set up like this?

    Follow Math Help Forum on Facebook and Google+

  11. #11
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Legend9591 View Post
    I'm confused by the second method of doing the equation.

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

    is [3/(x+1) - 1/(x-2)] right?

    But how did you 3(x-2) - 1(x+1) when the equation would be:

    3 1
    --- - ----
    x+1 x-2
    Notice that in the first method 3(x-2) - 1(x+1) was the numerator of the combined fraction. essentially the second method wipes out all denominators so you don't have to work with fractions anymore but just a linear equation. Here's what happens:

    we have 3/(x + 1) - 1/(x - 2) = 1/(x^2 - x - 2), and we also know (x^2 - x - 2) = (x + 1)(x - 2), so we can write the equation as:

    3/(x + 1) - 1/(x - 2) = 1/(x + 1)(x - 2)................got that part, right?

    okay, now we notice that the denominator on the right has factors that appear in the denominators on the left. So we go: ok, if i can wipe out the denominator on the right, it will also wipe out those on the left, so we multiply through by the denominator on the right:

    (x + 1)(x - 2)[ 3/(x + 1) - 1/(x - 2)] = (x + 1)(x - 2)[1/(x + 1)(x - 2)] .....still with me?

    so expanding the brackets we get:

    (x + 1)(x - 2)(3/(x + 1)) - (x + 1)(x - 2)(1/(x - 2)) = (x + 1)(x - 2) * 1/(x + 1)(x - 2)

    We immediately see that the denominator on the right cancels with what we multiply by.

    We also see that the (x + 1) of what we multiplied by cancels the (x + 1) in the denominator of 3/(x+1). so we are left with (x - 2)*3. similarly, the (x - 2) cancels the denominator of -1/(x - 2), so we are left with (x + 1)* (-1)

    so the equation becomes:
    3(x - 2) - (x + 1) = 1
    and we just solve for x from there.

    got it?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Legend9591 View Post
    I'm confused by the second method of doing the equation.

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

    is [3/(x+1) - 1/(x-2)] right?

    But how did you 3(x-2) - 1(x+1) when the equation would be:

    Would the equation be set up like this?

    yes, that is how the equation is set up. then you know you have to multiply each term in the bracket by what's on the outside. so the (x + 1)'s cancel in the first term and the (x - 2)'s cancel in the second term, and we are left with 3(x - 2) and -(x + 1).

    the denominator of the second fraction in brackets is x - 2
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Newbie
    Joined
    Mar 2007
    Posts
    12
    Okay I understand exactly what you mean now. I did it out on paper it makes perfect sense.

    Wow thank you... This helps a lot.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Legend9591 View Post
    Okay I understand exactly what you mean now. I did it out on paper it makes perfect sense.

    Wow thank you... This helps a lot.
    good! try a few more problems to make sure that you get it. using both methods to do each would be good
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Newbie
    Joined
    Mar 2007
    Posts
    12
    Okay I have one more question... When your factoring I always seem to forget what signs I need to use.

    x-x-x = (x- )(x+ )
    x-x+x = (x? )(x? )
    x+x-x = (x? )(x? )
    x+x+x = (x+ )(x+ )
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Rational equations?
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: December 11th 2010, 02:20 PM
  2. Rational Equations
    Posted in the Algebra Forum
    Replies: 2
    Last Post: May 20th 2010, 08:29 PM
  3. pre-alg rational equations
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 20th 2010, 07:28 AM
  4. Rational equations?
    Posted in the Algebra Forum
    Replies: 5
    Last Post: January 14th 2010, 02:43 PM
  5. rational equations
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: January 19th 2008, 06:24 PM

Search Tags


/mathhelpforum @mathhelpforum