Results 1 to 4 of 4

Thread: geometry

  1. #1
    Member
    Joined
    Mar 2006
    Posts
    82

    geometry

    plz help

    a, b, c, and d are real numbers and a > c, b> d, a/b = c/d = r
    Prove that a-c / b-d = r
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,152
    Thanks
    731
    Awards
    1
    Quote Originally Posted by jenjen View Post
    plz help

    a, b, c, and d are real numbers and a > c, b> d, a/b = c/d = r
    Prove that a-c / b-d = r
    Please use parenthesis. What you wrote is
    $\displaystyle a - \frac{c}{b} - d$

    To write it as you intended write (a - c)/(b - d) = r.

    $\displaystyle \frac{a}{b} = r$ means that $\displaystyle a = br$

    $\displaystyle \frac{c}{d} = r$ means that $\displaystyle c = dr$

    So
    $\displaystyle \frac{a - c}{b - d} = \frac{br - dr}{b - d} = \frac{r(b - d)}{b - d} = r$

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, jenjen!

    What have you tried?
    . . It's fairly straight-forward.


    $\displaystyle a,\,b,\,c,\,d$ are real numbers, and: .$\displaystyle a > c,\;\;b> d,\;\;\frac{a}{b}\,=\,\frac{c}{d}\,= \,r$

    Prove that: .$\displaystyle \frac{a-c}{b-d}\:=\:r$
    From $\displaystyle \frac{a}{b} \:=\:r$, we have: .$\displaystyle a \:=\:br$ .[1]

    From $\displaystyle \frac{c}{d} \:=\:r$, we have: .$\displaystyle c \:=\:dr$ .[2]


    Subtract [2] from [1]: .$\displaystyle a - c \:=\:br - dr\quad\Rightarrow\quad a-c\:=\:(b-d)r$


    Therefore: .$\displaystyle \frac{a - c}{b-d} \:=\:r$

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Mar 2006
    Posts
    82
    To Topsquark:

    Hi, thank you so much Topsquark. You save my life!


    To Soroban: I agreed with you. It is straightforward but then sometimes I can't see through simple problems. I often make math problems more complicated than it is and fool myself.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Finite Geometry: Young's Geometry
    Posted in the Advanced Math Topics Forum
    Replies: 0
    Last Post: Sep 15th 2010, 07:20 AM
  2. Geometry (1)
    Posted in the Geometry Forum
    Replies: 2
    Last Post: Nov 7th 2009, 06:44 PM
  3. geometry (2)
    Posted in the Geometry Forum
    Replies: 4
    Last Post: Nov 7th 2009, 06:41 PM
  4. geometry (3)
    Posted in the Geometry Forum
    Replies: 1
    Last Post: Nov 7th 2009, 07:22 AM
  5. Modern Geometry: Taxicab Geometry Problems
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: Mar 30th 2009, 07:32 PM

Search Tags


/mathhelpforum @mathhelpforum