An Equation With Integer Solutions?

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[@Qermaq]

(0, 0) = oh-oh! You broke math, man.

[@armacham]

cross-multiply and put everything on the same side to get 3xy - 4x - 6y = 0 multiply both sides by 3 so there's a square number in front of the xy 9xy - 12x - 18y = 0 we want to factor it into something like this: (3x - a)(3y - b) cross-multiply to get: 9xy - 3bx - 3ay + ab -3bx should be -12x so b = 4 -3ay should be -18y so a = 6 that gives us: (3x - 6)(3y - 4) = 9xy - 3bx - 3ay + 24 go back to the equation we have, and add 24 to both sides: 9xy - 12x - 18y = 0 9xy - 12x - 18y + 24= 24 (3x - 6)(3y - 4) = 24 (x-2)(3y - 4) = 8 on the left, the contents of parenthesis must be integers. And 8 doesn't have very many factor pairs: 1, 8 2, 4 4, 2 8, 1 -1, -8 -2, -4 -4, -2 -8, -1 from here it's easy to test all 8 possibilities where the first number is x-2 and the second number is 3y-4 to look for integer pairs x,y which are your solutions except of course 0,0

[@erikroberts8307]

(0, 0) is NOT a valid solution because if you plug these numbers into the original equation, you will be dividing by zero.

[@user-yt198]

6:49 He didn't take (0, 0) to final set of solutions.

[@wilsonli5642]

@@user-yt198 Yes but he didn't explain why he rejected it from the final set

[@SrisailamNavuluri]

1/2x + 1/3y =1/4 LCM(2,3,4)=12 Multiply both sides by 12. 6/x+4/y=12/4=3=1+2=2+1 x=6,y=2 or x=3,y=4.