Results 1 to 9 of 9

Math Help - Simple solving for X..

  1. #1
    Banned
    Joined
    Sep 2009
    Posts
    8

    Cool Simple solving for X..

    Its been a while since i took a math course, and i completely forgot how to solve for X in this equation.... y=4x-1/2x+3. I looked it up and found this from yahoo answers...

    1. (x) = (4x-1) / (2x+3)
    2 y = (4x-1) / (2x+3)
    3. y(2x+3) = 4x-1
    4. 2xy+3y = 4x-1
    5. 2xy - 4x = -3y-1
    6. 2x(y-2) = -3y-1
    7. x =(-3y-1)/[2(x-2)]
    8. f^-1(y) = (-3y-1)/[2(x-2)]


    ...my question is in step 5 above what did he do to get 2xy-4x from 2xy-3y, and also the procedure from steps 5 to 6 and 6 to 7.

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Sep 2009
    From
    Minnesota
    Posts
    80
    Thanks
    8
    In 4 -> 5, he subtracted 3y from both sides. Then he subtracted 4x from both sides. Notice that the other side changed as well. In 5 -> 6, he factored by grouping. If you still don't get it after looking at the link I can explain it to you but I think it goes over the topic pretty well. In 6 -> 7, he divided both sides by (y-2) to get 2x = (-3y-1)/(y-2) and then divided both sides by 2 to get just x on the left side. The author made a slight mistake when he wrote [2(x-2)] when is should be [2(y-2)]. Were you trying to solve for x or find the inverse function?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by evankimia View Post
    Its been a while since i took a math course, and i completely forgot how to solve for X in this equation.... y=4x-1/2x+3. I looked it up and found this from yahoo answers...

    1. (x) = (4x-1) / (2x+3)
    2 y = (4x-1) / (2x+3)
    3. y(2x+3) = 4x-1
    4. 2xy+3y = 4x-1
    5. 2xy - 4x = -3y-1
    6. 2x(y-2) = -3y-1
    7. x =(-3y-1)/[2(x-2)]
    8. f^-1(y) = (-3y-1)/[2(x-2)]


    ...my question is in step 5 above what did he do to get 2xy-4x from 2xy-3y, and also the procedure from steps 5 to 6 and 6 to 7.

    Thanks.
    4. 2xy+3y = 4x-1

    5. 2xy - 4x = -3y-1

    6. 2x(y-2) = -3y-1

    7. x = \frac{-3y-1}{2(x-2)}

    4-5: He collected the x terms on one side and the non x terms on the other side in order to factorise in step 6. This entailed taking 4x from both sides and taking 3y from both sides.

    5-6: He factored out a common term on the left which is 2x since both 2 and x are factors.

    6-7: He should have divided through by 2(y-2) which would give x = -\frac{3y+1}{2(y-2)}. That is what makes x the subject.

    Step 8 and parts of step 7 introduce the idea of finding the inverse function, is this what you wanted to do?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Sep 2009
    Posts
    8
    Yeah This Is For Solving For An Inverse Equation
    Follow Math Help Forum on Facebook and Google+

  5. #5
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by evankimia View Post
    Yeah This Is For Solving For An Inverse Equation
    For inverse functions we set the equation solved for x and replace that x with f^{-1}(x) and y with x.

    <br />
f^{-1}(x) = -\frac{3x+1}{2(x-2)}<br />

    The domain is x \in \mathbb{R} \: \: , \: \: x \neq 2

    The range is anything that satisfies the domain in f(x). x \in \mathbb{R} \: \: , \: \: x \neq -\frac{3}{2}

    edit: see the attached graph of f(x) (blue), g(x) = f^{-1}(x) (green) and h(x) = x (red). Since g(x) and f(x) are reflections in the line y=x they are inverse
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Banned
    Joined
    Sep 2009
    Posts
    8
    Thank you so much...learn something new every day. Just curious, what program did you use to make that function? -Evan
    Follow Math Help Forum on Facebook and Google+

  7. #7
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    I use KMPlot which is part of the KDE desktop usually associated with GNU/Linux

    It's licened under the GPL but the homepage is down at the moment :|
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Banned
    Joined
    Sep 2009
    Posts
    8
    I will check their homepage in a bit to see if its back up. In the meantime... heres another example.....

    Find the inverse of f(x)= (x+4)/(x-3) Use this to determine the domain and range of f and f-1.

    1. f(x)= (x+4)/(x-3) = y=(x+4)/(x-3)
    2. y(x-3)=x+4
    3. 1yx-3y=x+4
    how would i get the x out of there? just divide it out normally on both sides so it would be -2y=x+4/x?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by evankimia View Post
    I will check their homepage in a bit to see if its back up. In the meantime... heres another example.....

    Find the inverse of f(x)= (x+4)/(x-3) Use this to determine the domain and range of f and f-1.

    1. f(x)= (x+4)/(x-3) = y=(x+4)/(x-3)
    2. y(x-3)=x+4
    3. 1yx-3y=x+4
    how would i get the x out of there? just divide it out normally on both sides so it would be -2y=x+4/x?
    yx - 3y = x + 4

    => yx - x = 4 + 3y => x(y - 1) = 4 + 3y => x = (4 + 3y)/(y - 1).

    So the inverse function is f^{-1} = \frac{4 + 3x}{x - 1}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Need help solving simple integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 29th 2012, 04:34 AM
  2. Simple equation solving
    Posted in the Algebra Forum
    Replies: 7
    Last Post: September 18th 2011, 11:20 AM
  3. Simple Algebra - solving for R
    Posted in the Algebra Forum
    Replies: 2
    Last Post: January 18th 2011, 12:29 PM
  4. Simple equation solving
    Posted in the Algebra Forum
    Replies: 1
    Last Post: September 13th 2009, 02:14 PM
  5. Solving this simple function
    Posted in the Algebra Forum
    Replies: 3
    Last Post: March 9th 2009, 02:52 PM

Search Tags


/mathhelpforum @mathhelpforum