A Rational Equation with Integer Solutions (3/x)+(5/y)=4

If you need to post a picture of your solution or idea:
#NumberTheoryProblems #DiophantineEquations
EXPLORE 😎:

PLAYLISTS 🎵 :

[manojsurya]

The factorisation step was really cool, well done 😃

[rssl]

Hello I discovered the formula for general case let the equation be as the following :
a/b + c/d = x
ad + bc = bdx
bdx-bc-ad=0
bd(x- c/d)- a (x- c/d)=ac/d
(bd+a) (x - c/d ) =ac/d
(bd+a ) ( dx-c ) = ac
So you could straight up go ahead and solve this equation
I made this by my self :D

[einrazer]

I enjoyed this and it helps me break a little bit of my mathematical rigidity in solving. I was struggling with the fact that we couldn't use product of zeros, but it helped me go back to my roots of just having multiple options to test and observe. I would live to see if there is a direct way to solve this without the need of guess and check methodology though.

[UTAngoid]

Love the way you explain these things so simply!

[arundhatimukherjee]

(x, y) = (2, 2), (-3, 1) or (1, 5)
Simplification and simon with a little bit of modular arithmetic🙂🙂

[riade-yet]

Thanks. When i am watching your direct live i not only understand maths but also english speaking. You speak english fast.

[gmutubeacct]

A different approach...
3+5x/y = 4x
implies either y = 5 or x=ky for some integer k
y=5 gives you solution (1, 5)
if x = ky, y = (3+5k)/4k
implies k has to be odd and of the form 4i+1 ( so that (3 + 5k) can divide 4), and 3 has to divide k
only choices are k = 1 and k = -3
those will lead to solutions (2, 2) and (-3, 1) respectively

[babangmalas]

I will watch one video every day on this channel thank you sir

[breakingmath]

An interesting case in which the algebraic transformations increase the solution set, that's why we need to, always, keep our eyes on the original problem...

[nelad]

So, it's always a/x+b/x=c -> (cx-a)(cy-b)=ab

[mattslupek]

Without watching the video, I’m gonna say x=1, and y=5. 3/1=3. 5/5=1. 3+1=4. You’re welcome.

[Muslim_]

You complicated it more than it deserve.
From the equation it self you can get those solutions easly

[domsau]

When you are a real programmer, you make brute force!
;-)

[mrbennettmaths]

I tried a similar question on my channel, using your method of solving. Thank you for the method!

[solved]

Your videos inspired me a lot. It is simple but cool. I like your way of teaching.

[leecherlarry]

my compi thought that this was very cute problem haha:
*Solve[3/x + 5/y == 4, {x, y}, Integers]*

[MathElite]

Ooh integer solutions nice
I was thinking of just solving for y in terms of x

[tgx]

3y=(4y-5)x
4y-5=k where k Is integer & 3y= kx. We have then y=(5+k)/4 & k=15/(4x-3), we have solution for 4x-3=5, 4x-3=1, 4x-3=-15

[Grassmpl]

We immediately see that there are no solutions for which x>2. In this case 5/y>=3 so y=1. Hence the sum is at least 5, contradiction.

If either of them is negative, then it has to be x since 3/x can never exceed 4. Also y has to be 1 or else 5/y cannot exceed 4. We directly solve for x=-3.
The cases for x=1, 2 can be checked separately.

[NE]

1. 1 and 5 (3/1=3, 5/5=1 and 3+1=4) popped into my head when I first saw this.
2. 2 and 2 (3/2 and 5/2 give 8/2=4).
That's all from me.