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Easy Fraction Problem (Forgot how to do it )

  1. Jan 21, 2007 #1
    Easy Fraction Problem (Forgot how to do it!!)

    I came across a problem that I forgot how to do, please help me out here. thanks!

    Solve for x: [tex] \frac{1}{x-1}+\frac{3}{x+3}=1 [/tex]

    My questions: I know the common denominator is (x-1)(x+3) but do you multiply this number on both sides of the equal sign or just the left?

    Can anyone please show me step by step how to solve this? For some reason I keep thinking of it as:

    [tex] \frac{x+3}{x-1}+\frac{x-1}{x+3}=(x-1)(x+3) [/tex]

    I know this is wrong, right?

    Because I think that the x-1 and x+3 cancel out when you multiply them. Can someone clear up my misunderstandings?? This reason is I haven't done fractions in a year lol.
     
  2. jcsd
  3. Jan 21, 2007 #2
    Also, is the inverse of y=1/(3x-2) --> y=(1+2x)/3x?
     
  4. Jan 21, 2007 #3

    cristo

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    You cannot just multiply a few terms of an equation by something; you must multiply the whole equation. If you are confused, then multiply the whole equation by (x-1)(x+3), write it out in full, and then cancel the terms.
     
  5. Jan 21, 2007 #4
    So you multiply both sides of the equation by (x-1)(x+3)? Wow, I can't believe this is gone out of my brain =P
     
  6. Jan 21, 2007 #5

    cristo

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    Yes, the fact that this is an equation means that the left hand side is equal to the right hand side. If you only multiply the left hand side by, say, z, then the left hand side no longer equals the right hand side (in general) and so it is no longer an equation!
     
  7. Jan 21, 2007 #6
    Oh I see.. Whatever you do to one side you must do to the other? You need to multiply the whole side of the equation right? So you can't do something to 1 fraction and not to the other?
     
  8. Jan 22, 2007 #7

    uart

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    Well you can independantly multiply them by unity. That is, you multiply the first fraction by (x+3)/(x+3) and then you muliply the second fraction by (x-1)/(x-1), the right hand side of the equation just stays as is. That's is how this type of problem would normally be done. At first glance it might seem as if you're doing something different to each term, but since it is just multipying by unity you're not.
     
  9. Jan 22, 2007 #8

    HallsofIvy

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    There are two ways of handling
    [tex]\frac{1}{x-1}+ \frac{3}{x+3}= 1[/tex]

    One way is to go ahead and add the fractions on the left side- to do that you need to get the common denominator- which is obviously (x-1)(x+3)

    Multiply both numerator and denominator of the first fraction by x+ 3 and multiply both numerator and denominator of the second fraction by x-1:
    [tex]\frac{1}{x-1}+ \frac{3}{x+3}= \frac{x+3}{(x-1)(x+3)}+ \frac{3(x-1)}{(x-1)(x+3)}[/tex]
    [tex]= \frac{x+ 3+ 3x- 3}{(x-1)(x+3)}= \frac{4x}{(x-1)(x+3)}[/tex]
    Because we have not done anything new to the left hand side, just added what was already there, we don't need to do anything to the right side. We have
    [tex]\frac{4x}{(x-1)(x+3)}= 1[/tex]
    Now we can simplify by multiplying both sides by (x-1)(x+3). (There was no (x-1)(x+3) multiplying on the left side so this is new- we have to do it to both sides.) We get
    [tex] 4x= (x-1)(x+3)[/tex]
    Now you can multiply out the right side and solve the resulting quadratic equation. (It factors easily.)

    The other method is to just go ahead and multiply both sides of the equation by (x-1)(x+3):
    [tex]\frac{1}{x- 1}(x-1)(x+3)+ \frac{3}{x+3}(x-1)(x+3)= 1(x-1)(x+3)[/tex]
    [tex]x+ 3+ 3(x-1)= (x-1)(x+ 3)[/tex]
    [tex]4x = (x-1)(x+3)[/tex]
    as before. Most people prefer the second method.

    As for the inverse function to
    [tex]y= \frac{1}{3x-2}[/tex]
    First swap x and y:
    [tex]x= \frac{1}{3y- 2}[/tex]
    That's what really changes from the function to its inverse. But because we prefer to write a function y= ..., no we need to solve for x.
    Multiply both sides by 3y- 2 to get rid of the fraction:
    [tex]x(3y- 2)= 1[/tex]
    Divide on both sides by x:
    [tex]3y- 2= \frac{1}{x}[/tex]
    Add 2 to both sides:
    [tex] 3y= 2+ \frac{1}{x}[/tex]
    and finally divide both sides by 3:
    [tex]y= \frac{2+ \frac{1}{x}}{3}[/tex]
    which we can simplify by multiplying both numerator and denominator of the fraction by x (that's the same as multiplying by x/x= 1 so nothing new is done- we don't need to do that to the left side).
    [tex]y= \frac{2x+ 1}{3x}[/tex]
    just as you said.
     
    Last edited: Jul 19, 2011
  10. Jan 22, 2007 #9
    Thanks for the post HallsofIvy. I think I got confused because most people were doing it the first way and I was like what?? Don't the x+3 and x+3 (denominator) cancel out when you multiply it? lol. Has this ever happened to anyone?? Do you forget fractions in college? I'm learning Physics right now and I can't believe I would get fractions confused.. especially adding/multiplying
     
  11. Jul 19, 2011 #10
    Re: Easy Fraction Problem (Forgot how to do it!!)

    1/x-1 + 3/x+3 = 1
    let's first simplify the first two parts:
    1(x+3)+3(x-1)/(x-1)(x+3)=1
    if we multiply and cancel out we get:
    4x/(x-1)(x+3)=1
    so, let's make cross multiplication and we get:
    4x=x^2+2x-3
    then,
    x^2-2x-3=0
    x+1=0
    x-3=0
    so,
    x=-1
    or
    x=3
     
  12. Jul 19, 2011 #11

    HallsofIvy

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    Re: Easy Fraction Problem (Forgot how to do it!!)

    They could if you wanted them to but the don't have to. It is the fact that they could cancel that means you really haven't changed the numbers. A lot of algebra, and, in fact arithmetic, is making thing that are the same look different. Back in elementary school, to add 1/2+ 1/3, you get common denominators by multiplying the numerator and denominators of both sides. (1/2)(3/3)= 3/6. Yes, you could cancel the threes- that's why the fraction 3/6 is the same as 1/2. (1/3)(2/2)= 2/6. Yes, you could cancel the twos- thats why the fraction 2/6 is the same as 1/3. But you don't have to. Instead you can write 1/2+ 1/3= 3/6+ 2/6= 5/6.

     
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