Results 1 to 8 of 8

Math Help - Simple Equation

  1. #1
    Newbie
    Joined
    May 2010
    Posts
    22
    Thanks
    1

    Simple Equation

    I was wondering if someone could show me the way to write this one up.

    The sum of a number and it's reciprocal is 2 1/6 . What is the number? (sounds like fraction)


    I got this far......
    x+y= 2 1/6
    x= 2 1/6-y
    y= 2 1/6-x

    Where do I go from here to find the answer?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,404
    Thanks
    1293
    Quote Originally Posted by Tren301 View Post
    I was wondering if someone could show me the way to write this one up.

    The sum of a number and it's reciprocal is 2 1/6 . What is the number? (sounds like fraction)


    I got this far......
    x+y= 2 1/6
    x= 2 1/6-y
    y= 2 1/6-x

    Where do I go from here to find the answer?
    \frac{x}{y} + \frac{y}{x} = 2\frac{1}{6}

    \frac{x^2}{xy} + \frac{y^2}{xy} = \frac{13}{6}

    \frac{x^2 + y^2}{xy} = \frac{13}{6}.


    Therefore x^2 + y^2 = 13 and xy = 6.

    Solve the equations simultaneously.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,682
    Thanks
    614
    Hello, Tren301!

    Another approach . . .


    The sum of a number and it's reciprocal is 2\frac{1}{6}.
    What is the number?

    Let x = the number,

    then \frac{1}{x} = its reciprocal.


    Their sum is 2\frac{1}{6}\!:\;\;x + \dfrac{1}{x} \:=\:\dfrac{13}{6}

    Multiply by 6x\!:\;\;6x^2 + 6 \:=\:13x \quad\Rightarrow\quad 6x^2 - 13x + 6 \;=\:0

    Factor: . (3x-2)(2x-3) \:=\: 0

    . . \begin{array}{cccccccc}3x-2\:=\:0 & \Rightarrow & x \:=\:\frac{2}{3} \\ \\[-3mm]<br />
2x-3 \:=\:0 & \Rightarrow & x \:=\:\frac{3}{2} \end{array}


    Therefore: . x \;=\;\frac{2}{3}\text{ or }\frac{3}{2}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    May 2010
    Posts
    22
    Thanks
    1
    Hi there....you know I was wondering if you could show or explain the treatment I need to do to xy=6 (as it's multiplied) so that I can use it as substitution in x2+y2=13.
    It's probably simple, but I don't know. I understand the above fraction work in getting the common denominator.

    Thankyou.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by Tren301 View Post
    I was wondering if someone could show me the way to write this one up.

    The sum of a number and it's reciprocal is 2 1/6 . What is the number? (sounds like fraction)


    I got this far......
    x+y= 2 1/6
    x= 2 1/6-y
    y= 2 1/6-x

    Where do I go from here to find the answer?
    You could also try...

    2+\frac{1}{6}=\frac{12}{6}+\frac{1}{6}>2

    Therefore one number must be >1

    \frac{a+b}{b}+\frac{b}{a+b}=\frac{(a+b)(a+b)+b^2}{  b(a+b)}=\frac{a^2+ab+ab+b^2+b^2}{b(a+b)}

    =\frac{a^2+ab+b^2+b(a+b)}{b(a+b)}

    \frac{13}{6}=\frac{6+7}{6}

    \Rightarrow\ b(a+b)=6,\ a^2+ab+b^2=7

    b(a+b)=6(1) or 3(2).. hence it's 3(2) so b=2 and a+b=3 so a=1.

    Therefore the numbers are

    \frac{a+b}{b}=\frac{3}{2},\ \frac{b}{a+b}=\frac{2}{3}
    Last edited by Archie Meade; July 29th 2010 at 05:38 AM. Reason: typo
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by Tren301 View Post
    Hi there....you know I was wondering if you could show or explain the treatment I need to do to xy=6 (as it's multiplied) so that I can use it as substitution in x2+y2=13.
    It's probably simple, but I don't know. I understand the above fraction work in getting the common denominator.

    Thankyou.
    Simplest is...we are looking for integers x and y, so you can just factor the 6 into 3(2).

    Or 13=1+12, 2+11, 3+10, 4+9....4 and 9 are squares.

    9+4=3^2+2^2

    You could do it purely in algebra with xy=6\ \Rightarrow\ y=\frac{6}{x}

    Now substitute this into the other to get x only..

    x^2+\frac{6^2}{x^2}=13

    Multiply both sides by x^2

    x^4+36=13x^2

    Both are equal, subtract to get zero, then factor

    x^4-13x^2+36=0

    \left(x^2-9\right)\left(x^2-4\right)=0

    then x^2=9,\ x^2=4

    etc
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    May 2010
    Posts
    22
    Thanks
    1
    Thanks Archi,
    I could follow your second example better. So x=3, or x=2. Is there a process for turning that into a fraction of 3/2 0r 2/3, or do we just use logic, to explain that it is their sum that equals to 2 1/6, so therefore that is it? I liked the logic of explaining the integers! With the first example I don't see how you got 3 lots of (a+b) in the numerator...(a+b)(a+b)b(a+b)over b(a+b). I thought it would have been (a+b)(a+b)b2 over b(a+b). Don't you multilpy the numerator and denominator, by the other fractions denominator to balance the fractions? I understand the rest of the example after that. But in the first place, how did you come up with (a+b)/b and b/(a+b). I understand the use of a/b and b/a. Pardon my ignorance, and thanks for your time.

    Tren301
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by Tren301 View Post
    Thanks Archi,
    I could follow your second example better. So x=3, or x=2. Is there a process for turning that into a fraction of 3/2 0r 2/3, or do we just use logic, to explain that it is their sum that equals to 2 1/6, so therefore that is it? I liked the logic of explaining the integers! With the first example I don't see how you got 3 lots of (a+b) in the numerator...(a+b)(a+b)b(a+b)over b(a+b). I thought it would have been (a+b)(a+b)b2 over b(a+b). Don't you multilpy the numerator and denominator, by the other fractions denominator to balance the fractions? I understand the rest of the example after that. But in the first place, how did you come up with (a+b)/b and b/(a+b). I understand the use of a/b and b/a. Pardon my ignorance, and thanks for your time.

    Tren301
    Hi Tren301,

    \frac{3}{2}+\frac{2}{3}=\left(\frac{3}{3}\right)\f  rac{3}{2}+\left(\frac{2}{2}\right)\frac{2}{3}

    =\frac{9}{6}+\frac{4}{6}=\frac{13}{6}

    The way I did it in the first post is the same as that.

    I used \frac{a+b}{b} as one of the numbers is bigger than 1, since \frac{1}{1}+\frac{1}{1}=2

    in other words, one of the numbers cannot be 1, so one is bigger than 1 and the other is smaller than 1.

    \frac{a+b}{b}+\frac{b}{a+b}=\left(\frac{a+b}{a+b}\  right)\frac{a+b}{b}+\left(\frac{b}{b}\right)\frac{  b}{a+b}

    Those two fractions now have a common denominator, so the numerators can be combined.

    There was a typo in that earlier post... now fixed.
    Last edited by Archie Meade; July 29th 2010 at 05:40 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Simple PDE equation
    Posted in the Differential Equations Forum
    Replies: 17
    Last Post: June 28th 2011, 06:13 AM
  2. Simple Equation Help
    Posted in the Algebra Forum
    Replies: 15
    Last Post: July 24th 2010, 06:07 AM
  3. simple log equation....
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: October 21st 2008, 04:33 PM
  4. Simple equation help
    Posted in the Algebra Forum
    Replies: 5
    Last Post: July 30th 2008, 07:28 AM
  5. A Very simple equation
    Posted in the Trigonometry Forum
    Replies: 9
    Last Post: March 9th 2006, 12:01 PM

Search Tags


/mathhelpforum @mathhelpforum