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  1. #1
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    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?
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  2. #2
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    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?
    $\displaystyle \frac{x}{y} + \frac{y}{x} = 2\frac{1}{6}$

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

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


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

    Solve the equations simultaneously.
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  3. #3
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    Hello, Tren301!

    Another approach . . .


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

    Let $\displaystyle x$ = the number,

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


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

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

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

    . . $\displaystyle \begin{array}{cccccccc}3x-2\:=\:0 & \Rightarrow & x \:=\:\frac{2}{3} \\ \\[-3mm]
    2x-3 \:=\:0 & \Rightarrow & x \:=\:\frac{3}{2} \end{array}$


    Therefore: . $\displaystyle x \;=\;\frac{2}{3}\text{ or }\frac{3}{2}$
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    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.
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    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...

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

    Therefore one number must be >1

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

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

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

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

    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

    $\displaystyle \frac{a+b}{b}=\frac{3}{2},\ \frac{b}{a+b}=\frac{2}{3}$
    Last edited by Archie Meade; Jul 29th 2010 at 05:38 AM. Reason: typo
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    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.

    $\displaystyle 9+4=3^2+2^2$

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

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

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

    Multiply both sides by $\displaystyle x^2$

    $\displaystyle x^4+36=13x^2$

    Both are equal, subtract to get zero, then factor

    $\displaystyle x^4-13x^2+36=0$

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

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

    etc
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    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
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  8. #8
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    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,

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

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

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

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

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

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

    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; Jul 29th 2010 at 05:40 AM.
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