Solve the equations:
4x^2-5xy+y^2=10,
3x^2-3xy-y^2=5.
.
.
.
ANSWER = {(1,-1),(-1,1),(3,2),(-3,-2)}

Question

Solve the equations:
4x^2-5xy+y^2=10,
3x^2-3xy-y^2=5.
.
.
.
ANSWER = {(1,-1),(-1,1),(3,2),(-3,-2)}

anonymous · Answer

the equations are so healthy what should i do with them

anonymous · Answer

hello? how should i start

hartnn · Answer

idk, this requires some thinking, i'll be back.....

anonymous · Answer

ok

anonymous · Answer

i am majin vegeta

anonymous · Answer

factoring........

anonymous · Answer

Try solving the simultaneous equation. :)

anonymous · Answer

how??

experimentx · Answer

each is that's equation of conic. that might represent from pair to two lines to hyperbola. http://en.wikipedia.org/wiki/Conic_section#Discriminant_classification one way is to solve for x in terms of y in 4x^2-5xy+y^2=10, and substitute in another equation ... which is quartic in y another method, reduct these equations to their standard form, it would be easy for them to solve after that.

anonymous · Answer

thats too hard i am in 11th grade

anonymous · Answer

@koli123able I remember you said 2 days ago that " I'm just a 10th grader" :0

anonymous · Answer

yeah did it by mistake.... nicely caught.. XD

experimentx · Answer

if you are 11 the grader you should be able to do factoring and quadratic equations moreover this is hyperhyperbola http://www.wolframalpha.com/input/?i=4x^2-5xy%2By^2%3D10%2C

experimentx · Answer

both are hyperbolas http://www.wolframalpha.com/input/?i=3x^2-3xy-y^2%3D5 do this 4x^2-5xy+y^2=10 <-- this is quadratic in x ... solve for x in terms of y. substitute this x in here 3x^2-3xy-y^2=5 remove all radical signs by squaring ... you will have quartic equation ... solve it by factoring.