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Math Help - Rational Equations

  1. #16
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Legend9591 View Post
    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+ )
    first of all, you can only have one x, so i replaced the second x with c, meaning a constant, b is also a constant

    x-bx-c = (x- )(x+ )
    x-bx+c = (x- )(x- )
    x+bx-c = (x+ )(x- )
    x+bx+c = (x+ )(x+ )

    Here are the rules:

    Multiplying two things of the same sign gives a positive sign for the result. that is, a positive number times a positive number gives a positive number and a negative number times a negative number also gives you a positive number

    Multiplying two things of opposite sign gives a negative sign for the result. that is, multiplying a positive number by a negative number (or vice versa) give a negative number.

    when factorizing, you want to make sure the two numbers you pick, when they multiply you get the constant and when you add them, you get the coefficient of x.

    so for
    x-bx-c
    we see the constant has a negative sign, so the two numbers we find must have opposite signs, so you get (x - *) and (x + *)

    for x-bx+c

    we notice the constant has a positive sign, so we need the same sign for both numbers we find. (x + ) and (x + ) wont work, since the coefficient of x is negative (if we add two positive numbers the result is positive--always). so we have to have (x - ) and (x - )

    for x+bx-c

    we need (x + ) and (x - ) since a negative times a positive gives a negative for the constant and a positive plus a negative gives a positive if the positive number is bigger for the coefficient of x

    for x+bx+c
    we need (x + ) and (x + ) since a positive times a positive is positive for the constant and a positive plus a positive is positive for the coefficient of x
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  2. #17
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    Okay I just started this new equation:

    x/(x+3) +1/(x-1) =4/(x+3)(x-1)

    and I followed all the steps and got x+3=4

    So should I subtract the 3 and get

    x=1

    and then take the sqaure root and get 1?
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  3. #18
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Legend9591 View Post
    Okay I just started this new equation:

    x/(x+3) +1/(x-1) =4/(x+3)(x-1)

    and I followed all the steps and got x+3=4

    So should I subtract the 3 and get

    x=1

    and then take the sqaure root and get 1?
    yes, however, your answer must be +/- 1. since (-1)^2 = (1)^2 = 1.

    so when you take square roots, or any other even root for that matter (such as 4th root, 16th root etc...), you must add a +/- to the answer
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  4. #19
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    Okay...

    I finished that equation and then I moved onto an equation later on that says

    1/(y-16) - 2/(y+4) = 2/(y-4)

    Same concept? I factor the y - 16 into (y-4)(y+4) and then get a common denominator and multiply by (y-4)(y+4) to eliminate the denominator?
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  5. #20
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Legend9591 View Post
    Okay...

    I finished that equation and then I moved onto an equation later on that says

    1/(y-16) - 2/(y+4) = 2/(y-4)

    Same concept? I factor the y - 16 into (y-4)(y+4) and then get a common denominator and multiply by (y-4)(y+4) to eliminate the denominator?
    yes, same concept, follow either method and you get the answer
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  6. #21
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    Hey I wanted to thank you for helping me out.
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  7. #22
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Legend9591 View Post
    Hey I wanted to thank you for helping me out.
    you're welcome, that's what i'm here for. so your good with these kinds of problems now right? if you get one on the test, you just do it in one step and write "don't insult my intelligence!" and just move on
    Last edited by Jhevon; March 4th 2007 at 05:42 PM.
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