Simple Equation

**Tren301** · June 29th 2010, 07:35 PM

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?

**Prove It** · June 29th 2010, 07:40 PM

Originally Posted by Tren301

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.

**Soroban** · June 29th 2010, 08:03 PM

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

**Tren301** · July 28th 2010, 10:12 AM

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.

**Archie Meade** · July 28th 2010, 11:31 AM

Originally Posted by Tren301

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

$\frac{a+b}{b}=\frac{3}{2},\ \frac{b}{a+b}=\frac{2}{3}$

**Archie Meade** · July 28th 2010, 11:41 AM

Originally Posted by Tren301

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

**Tren301** · July 29th 2010, 03:42 AM

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

**Archie Meade** · July 29th 2010, 04:02 AM

Originally Posted by Tren301

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.

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