3x + 5y = 2xy - 1. Find all possible answers of x and y.(They must be positive)?

Question

3x + 5y = 2xy - 1. Find all possible answers of x and y.(They must be positive).

I have found two answers. I don't think there is anymore...i have my way of doing it and i can SHOW it as i can reach 100 trial and error with my proof but it doesnt prove that x and y cannot be higher...can anyone help?

Answer 1

G'day,

Okay so the problem you have been set with is

3x + 5y = 2xy -1

where

x>=0 y>=0

This system of equations is known as an undertermined system of equations in that for the two unknowns x, y there is only one governing equation and subsequently so unique solution can be obtained, this is not necesarily a bad thing, however more than like infinite solutions will exist

for simplicity, reaarrange your equation to the form y =

3x + 5y = 2xy - 1

2xy - 5y = 3x + 1

y(2x - 5) = 3x +1

y = (3x + 1)/(2x-5)

Here we will let x be some free variable t and solve for t so that the positivity constrainsts are met,

Thus x = t

not x>=0 --> t>=0

Hence we have our first condition on t

now consider y

y = y = (3x + 1)/(2x-5) --> y = (3t + 1)/(2t-5)

for y to be positive two situations must be taken into account

(i)   (3t-1) > 0 and (2t - 5) > 0 (i.e.) positive x positive = positive

and

(ii)   (3t-1) < 0 and (2t - 5) < 0 (i.e) negative x negative = positive

for (i)

(3t-1) > 0 and (2t - 5) > 0

solving yields

t > 1/3 and t>2/5

Recall our constraint on x being t > 0

Thus we now have three constraints on x,y being positive namely

t>0, t> 1/3, t> 2/5

Hence for all three to be satisfied

t> 2/5 (new constraint on t)

for (ii)

(3t-1) < 0 and (2t - 5) < 0

solving yields

t < 1/3 and t < 2/5

Recall our constraint on x being t > 0

Hence for all three to be satisfied t < 1/3

Combining our two constraints on t we have

t>0, t < 1/3,  t > 2/5.

Hence our solution (x,y) = (t, (3x + 1)/(2x-5) )

There exists infinite solutions for the domain for which

0 < t < 1/3 and t>2/5

Hope this helps,

David

Answer 2

I think you mean integers else infinite number of solutions

let is get y in terms of x

3x + 1 = 2xy - 5y

= y(2x-5)

so y = (3x+1)/(2x-5) = 1 + (x+6)/(2x-5)

x+6 = 0 means x is -ve

x-6 = 0 means x is not integer ruled out

x+6 >=  2x - 5 x <= 11 if x > 11 we shall have y <=0

1 + (x+6)/(2x-5)

x= = 1 y = not integer

x = 2 y = 1 + 8/(-1) -ve

x = 3 y = 1 + 9/1 = 10 one solution

x = 4 y = not integer

x = 6 y = not integer

x =7 y not integer

x =8 y = not integer

x = 9 y not integer

x = 10 y not integer

x = 11 y = 2

so only solutions are (11,2) and (3,10)

edit

In eseponse to

@ maths_kp: so y = (3x+1)/(2x-5) = 1 + (x+6)/(2x-5)

how did you get y=1+(x+6)/(2x-5)?

and what is '-ve' I apologise but can you elaborate a little more?

(3x+1) = (2x-5 + x + 6)

so (3x+1)/ (2x-5) + (x+6)/ (2x-5)

basically I have separated to integral and fractional part in algebra(aritmetically send part may be integer

and secondly -ve means negative or < 0