#4**+2 **

\(\frac{2}{a} - \frac{3}{b} = \frac{1}{5}\\\)

**Lets look at Partial Mathemticians answers: **

**They are (10, 4), (15, 5), (35, 7), (60, 8), and (135, 9).**

(10, 4) 2/10-3/4 not equal 1/5 doesn't work,

(15, 5), 2/15 -3/5 =-7/15 doesn't work

(35, 7), 2/35-3/7 = -13/35 doesn't work

(60, 8), 2/60 - 3/8 = -41/120 doesn't work

and (135, 9). 2/135 -3/9 = -43/135 doesn't work

None of his answers work, maybe the a and b are the wrong way around. I have not looked at that.

**No one should asssume that any answer is correct. It is up to the question asker to check.**

\(\frac{2}{a} - \frac{3}{b} = \frac{1}{5}\\ 5ab(\frac{2}{a} - \frac{3}{b})= 5ab(\frac{1}{5})\\ 10b-15a=ab\\ \)

.Melody Dec 21, 2018

#8**+1 **

Well Rom, I agree with you and Melody, but if W/A and Mathematica 11 Home edition for that matter, don't give ANY explanations as to how they derive these numbers, then all I can think of is that they must use iteration, or trial and error essentially, to get these numbers!!. Since I have Newton-Raphson iteration and interpolation method programmed into my computer, I can essentially do the same thing they do.

Sorry, that is the only explanation I have.

Guest Dec 21, 2018

#11**0 **

**The new generation …**

I suspected from the beginning that Chimp Loki used recombinant DNA from the **four (4) Mr. BBs** to create a *mathematician*. The experiment was a **partial** success –Loki created a **PartialMathematician**

For those unfamiliar with the act III scenes of Web2.0calc’s Opera *Camelot*__(__Camel-Lot,)

here’s a **partial** synopsis of the *BB’s*:

https://web2.0calc.com/questions/kim-has-10-identical-lamps-and-3-identical-tables#r11

**Partial** list of posts with more detailed summaries on the pervasive phage of the BBs:

https://web2.0calc.com/questions/a-bag-contains-different-colored-beads#r10

https://web2.0calc.com/questions/help_77545#r5

https://web2.0calc.com/questions/hallppp-3-questions#r4

The saga continues …

GA

GingerAle
Dec 21, 2018

#13**0 **

I think something like this has more of a number theory type solution.

Chinese remainder theorem or something.

I will tinker with it and see if anything pops out.

Rom
Dec 21, 2018

#14**0 **

Hi Rom,

Thanks, but I hope you do not think it was a request from me.

This question has NOT captured my imagination.

I was just trying to make the asker and other answerers think a little more.

If you are looking at it because it interests you personally then I appologise for butting in :)

Melody
Dec 22, 2018

#5**+1 **

There are 5 pairs of POSITIVE integers that satisfy the equation. See here:

**https://www.wolframalpha.com/input/?i=2%2Fa+-+3%2Fb+%3D1%2F5,+iterate+for+a,+b**

Guest Dec 21, 2018

#9**0 **

If \(\displaystyle \frac{2}{a}-\frac{3}{b}=\frac{1}{5},\)

then

\(\displaystyle 10b-15a=ab,\)

so

\(\displaystyle b =\frac{15a}{10-a}.\)

\(\displaystyle \text{Both }a\text{ and } b\text{ are to be positive integers,}\)

so there are just nine possible values for \(\displaystyle a ,\;\;\; 1,2,\dots,9.\)

Just check each one in turn.

Guest Dec 21, 2018