Results 1 to 4 of 4

Math Help - Solving for x

  1. #1
    Newbie
    Joined
    Nov 2008
    Posts
    17

    Solving for x

    Can someone help me solve for x? I tried many different ways, but I never got the answer so can someone show me the steps so I can understand what I did wrong?

    1.
    y = 1 - \frac {1}{1-x}
    The answer should be x = \frac {y}{y-1}, but I don't know how to get that answer...


    2.
     y = 1 + \frac {1}{1+ \frac {1}{1-x}}
    The same for this one, i don't know how to get the answer  x= \frac {2y -3}{y-2}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    I have

     y = 1 - \frac {1}{1-x}

     \frac {1}{1-x}=1-y

     1=(1-y)(1-x)

     \frac {1}{1-y}=1-x

     \frac {1}{1-y}-1=-x

     -\frac {1}{1-y}+1=x

     1-\frac {1}{1-y}=x

     x=1-\frac {1}{1-y}

     x=\frac{1-y}{1-y}-\frac {1}{1-y}

     x=\frac {1-y-1}{1-y}

     x=\frac {-y}{1-y}

     x=\frac {y}{y-1}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2008
    Posts
    17
    2. <br />
y = 1 + \frac {1}{1+ \frac {1}{1-x}}<br />


    For the second one, is the next step:
    y-1 = \frac {1}{1+ \frac {1}{1-x}}

    then:
    (y-1)(1+ \frac {1}{1-x}) = \frac {1}{1+ \frac {1}{1-x}}(1+ \frac {1}{1-x})
    (y-1)(1+ \frac {1}{1-x}) = 1

    I don't know if I am right, but this is where I'm stuck at...
    can anyone help me with the next steps to get the answer x= \frac {2y -3}{y-2}?
    Last edited by ninjuhtime; November 4th 2009 at 10:25 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,693
    Thanks
    1466
    Quote Originally Posted by ninjuhtime View Post
    2. <br />
y = 1 + \frac {1}{1+ \frac {1}{1-x}}<br />


    For the second one, is the next step:
    y-1 = \frac {1}{1+ \frac {1}{1-x}}
    Yes!

    then:
    (y-1)(1+ \frac {1}{1-x}) = \frac {1}{1+ \frac {1}{1-x}}(1+ \frac {1}{1-x})
    (y-1)(1+ \frac {1}{1-x}) = 1
    You can do that but I think it is simpler first to clear the fraction on the right side by multiplying both numerator and denominator by 1-x:
    y-1= \frac {1(1-x)}{(1+ \frac {1}{1-x})(1-x)}= \frac{1-x}{(1-x)+ 1} = \frac{1-x}{2-x}

    I don't know if I am right, but this is where I'm stuck at...
    can anyone help me with the next steps to get the answer x= \frac {2y -3}{y-2}?
    From y- 1= \frac{1-x}{2-x}, you can multiply both sides by 2- x to get (2-x)(y-1)= 1-x so 2y-2- (y-1)x= 1- x. Add x to both sides and subtract 2y-2 from both sides: x-(y-1)x= 1+2- 2y or (2-y)x= 3- 2y.

    Finally, divide both sides by 2-y.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. solving for x
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: November 7th 2009, 04:13 PM
  2. help solving for y
    Posted in the Algebra Forum
    Replies: 2
    Last Post: November 1st 2009, 10:25 AM
  3. help me in solving this
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: December 13th 2008, 01:51 PM
  4. Solving x'Px = v
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: December 11th 2008, 03:21 PM
  5. Replies: 3
    Last Post: October 11th 2006, 09:15 PM

Search Tags


/mathhelpforum @mathhelpforum