+118690
\(\frac{2}{a} - \frac{3}{b} = \frac{1}{5}\\\)
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.
\(\frac{2}{a} - \frac{3}{b} = \frac{1}{5}\\ 5ab(\frac{2}{a} - \frac{3}{b})= 5ab(\frac{1}{5})\\ 10b-15a=ab\\ \)
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.
+2539 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
+6250 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.
+118690 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 :)
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
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.
