Question

I know how can x^2+xy+y^2=0 be solved but how can I solve 3x^2+8xy-3y^2=2012 ? there is 2012 instead of 0

Answer 1

there are 2 equations ?

Answer 2

no, only 3x^2+8xy-3y^2=2012

Answer 3

with x, y were integer numbers right ?

Answer 4

yes

Answer 5

ok, first make factoring it will giving us
(3x-y)(x+3y) = 2012 right ?

Answer 6

yes. But how did you realize that? :)

Answer 7

it's simple, try and errors :p

Answer 8

i meant, with try rearranged to get the factor of them

Answer 9

ok ok. so then?

Answer 10

u must find the factors of 2012 too :)

Answer 11

2012=2*3*337

Answer 12

2012 cant divided by 3 :)

Answer 13

ah, I was doing it on my mind :D and I confused 1006 with 1011...

Answer 14

ok, can you continue and solve it? :)

Answer 15

i can, but i must let u doing together :p

Answer 16

remember, the rule's from "Openstudy" :p

Answer 17

I'm new here :D

Answer 18

but, im not sure u cant find the factors of 2012, right ? :)

Answer 19

2^2*503

Answer 20

yes, nice.. :)

Answer 21

but it should be 2012 = +- {1,2,4,503,1006,2012} right ?

Answer 22

yes, I just found prime factors :D

Answer 23

ok, now any posibles that can we doing from there, for an example :
(3x-y)(x+3y) = 1 * 2012 , it means
3x-y=1 and x+3y=2012, with elimination and subtitution method, we can find the solution of them, it getting us x=201.5 and y=... but, exactly x is not an integer number, so not satisfies from the 1st posible equation, can u find the others ?

Answer 24

ok, I got it. thanks ;)

Answer 25

ok, glad to hear u can understand this problem.. :)
very welcome

Answer 26

but, as compare, u can look the answer in wolfram :
http://www.wolframalpha.com/input/?i=3x^2%2B8xy-3y^2%3D2012