Page 1 of 2 12 LastLast
Results 1 to 15 of 19

Math Help - Transposition of formulae question

  1. #1
    Junior Member
    Joined
    May 2011
    Posts
    37

    Transposition of formulae question

    Hi guys I'm having difficulties approaching some of these problems, pointing me toward the right direction and checking my answers would be much appreciated

    Question:

    Rearrange each of the following equations making x the subject:

    1.) w= y/ x + c
    = x = yc-wc/w-c

    2.) y^3= (xz)^3
    = x= sqroot y^3-z^3

    3.) zx= yx + w
    = x= w/z-y

    4.) v= 8^3/2......No idea about this one

    In the likely event that I've got all/any wrong I can provide the steps which got me into this mess on request.....thanks in advance guys
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Quacky's Avatar
    Joined
    Nov 2009
    From
    Windsor, South-East England
    Posts
    901
    Quote Originally Posted by komodo View Post
    Hi guys I'm having difficulties approaching some of these problems, pointing me toward the right direction and checking my answers would be much appreciated

    Question:

    Rearrange each of the following equations making x the subject:

    1.) w= y/ x + c
    = x = yc-wc/w-c

    2.) y^3= (xz)^3
    = x= sqroot y^3-z^3

    3.) zx= yx + w
    = x= w/z-y

    4.) v= 8^3/2......No idea about this one

    In the likely event that I've got all/any wrong I can provide the steps which got me into this mess on request.....thanks in advance guys
    This is very unclear. Firstly, 4) cannot be rearranged as it is - there is no x - did you leave a typo? Or did you just want to solve 8^3/2?

    As for the others:

    1)w=\frac{y}{x}+c
    Subtract c from both sides:
    w-c=\frac{y}{x}
    Multiply both sides by x:
    x(w-c)=y
    Divide both sides by the factored term:
    x=\frac{y}{w-c}

    2) y^3=(xz)^3
    Take the cube root of both sides:
    y=xz
    Divide both sides by z:
    x=\frac{y}{z}

    3) You have the correct solution (I believe), but your notation is incorrect. Think about brackets, and the order of operations.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    May 2011
    Posts
    37
    Sorry 4.) should be v=8x^3/2

    I'm guessing for 3 there should be a brackets around z-y?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Quacky's Avatar
    Joined
    Nov 2009
    From
    Windsor, South-East England
    Posts
    901
    Quote Originally Posted by komodo View Post
    Sorry 4.) should be v=8x^3/2

    I'm guessing for 3 there should be a brackets around z-y?
    Yeah. Do you think you can have a go at 4) by yourself, using the above concepts? What is the inverse of dividing by two, multiplying by eight and cubing something? Take it one step at a time.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    May 2011
    Posts
    37
    Quote Originally Posted by Quacky View Post
    Yeah. Do you think you can have a go at 4) by yourself, using the above concepts? What is the inverse of dividing by two, multiplying by eight and cubing something? Take it one step at a time.
    Going step by step this is what I got:

    v=8x^3/2

    = v/8 = x^3/2

    = 2.v/8 = x^3

    x= cuberoot 2.v/8

    Close enough???
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member Quacky's Avatar
    Joined
    Nov 2009
    From
    Windsor, South-East England
    Posts
    901
    Quote Originally Posted by komodo View Post
    Going step by step this is what I got:

    v=8x^3/2

    = v/8 = x^3/2

    = 2.v/8 = x^3

    x= Squareroot 2.v/8

    Close enough???
    Yeah, except you want to take the cube root at the last stage, not the square root. And remember that \frac{2v}{8}=\frac{v}{4}

    Edit: I see that you've corrected this yourself while I was replying.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    May 2011
    Posts
    37
    Quote Originally Posted by Quacky View Post
    Yeah, except you want to take the cube root at the last stage, not the square root. and remember that \frac{2v}{8}=\frac{v}{4}
    Thats great thanks mate

    By the way the 1st question should have been w= y/ (x+b) so in other words y divided by both of them not just y/x +c.....my fault I should have written it down more clearly.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member Quacky's Avatar
    Joined
    Nov 2009
    From
    Windsor, South-East England
    Posts
    901
    Quote Originally Posted by komodo View Post
    Thats great thanks mate

    By the way the 1st question should have been w= y/ (x+b) so in other words y divided by both of them not just y/x +c.....my fault I should have written it down more clearly.
    Ok, start by multiplying both sides of the equation by (x+b) to give:
    w(x+b)=y

    Can you take it from here? Expand the brackets, isolate the terms involving x, and from there, finishing should be quite simple.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    May 2011
    Posts
    37
    Quote Originally Posted by Quacky View Post
    Ok, start by multiplying both sides of the equation by (x+b) to give:
    w(x+b)=y

    Can you take it from here? Expand the brackets, isolate the terms involving x, and from there, finishing should be quite simple.
    w(x+b)=y

    wx+wb=y

    wx= y-wb

    x= y-wb / w

    Is this correct?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member Quacky's Avatar
    Joined
    Nov 2009
    From
    Windsor, South-East England
    Posts
    901
    Quote Originally Posted by komodo View Post
    w(x+b)=y

    wx+wb=y

    wx= y-wb

    x= y-wb / w

    Is this correct?
    Yep.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member
    Joined
    May 2011
    Posts
    37
    Quote Originally Posted by Quacky View Post
    Yep.
    Thanks for your help once again much appreciated
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,835
    Thanks
    1604
    Quote Originally Posted by komodo View Post
    w(x+b)=y

    wx+wb=y

    wx= y-wb

    x= y-wb / w

    Is this correct?
    No it's not. You MUST use the brackets when you intend for subtraction to happen before division.

    It should be (y - wb)/w
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Super Member
    Joined
    Nov 2007
    From
    Trumbull Ct
    Posts
    918
    Thanks
    27
    Hi komodo,
    solve for x v =8x^3/2
    rearrange x^3/2 =v/8
    raise each equality to the 2/3 power ((x)^3/2)^2/3= (v^2/3 )/8^2/3
    simplify x=( v^2/3)/4
    Follow Math Help Forum on Facebook and Google+

  14. #14
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,835
    Thanks
    1604
    This is the perfect example of why you need to use brackets where they're needed.

    We can't tell if it's \displaystyle v = \frac{8x^3}{2} or \displaystyle v = 8x^{\frac{3}{2}}
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Junior Member
    Joined
    May 2011
    Posts
    37
    Quote Originally Posted by Prove It View Post
    This is the perfect example of why you need to use brackets where they're needed.

    We can't tell if it's \displaystyle v = \frac{8x^3}{2} or \displaystyle v = 8x^{\frac{3}{2}}
    It should have been \displaystyle v = 8x^{\frac{3}{2}}


    Quote Originally Posted by Prove It View Post
    No it's not. You MUST use the brackets when you intend for subtraction to happen before division.

    It should be (y - wb)/w
    Thanks

    Quote Originally Posted by bjhopper View Post
    Hi komodo,
    solve for x v =8x^3/2
    rearrange x^3/2 =v/8
    raise each equality to the 2/3 power ((x)^3/2)^2/3= (v^2/3 )/8^2/3
    simplify x=( v^2/3)/4
    Hi bjhopper

    Is x true for both x=(v^2/3) /4 and x=cuberoot v / 4 or have I confused everyone by not writting it down correctly?


    Thanks guys
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: June 6th 2011, 02:27 AM
  2. Transposition of formulae
    Posted in the Algebra Forum
    Replies: 4
    Last Post: November 2nd 2009, 03:54 AM
  3. Transposition of Formulae
    Posted in the Algebra Forum
    Replies: 4
    Last Post: October 25th 2009, 11:07 AM
  4. Transposition question
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 7th 2009, 06:07 PM
  5. Transposition Question
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: August 8th 2008, 02:58 PM

Search Tags


/mathhelpforum @mathhelpforum