SOLUTION: Solve: 4x^2-3y^2=24 2x^2+3y^2=30 I know x=3 and y=2 but I just don't know in what form to put these in, i.e. (x,y)

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: Solve: 4x^2-3y^2=24 2x^2+3y^2=30 I know x=3 and y=2 but I just don't know in what form to put these in, i.e. (x,y)      Log On


   

Question 79336: Solve:
4x^2-3y^2=24
2x^2+3y^2=30
I know x=3 and y=2 but I just don't know in what form to put these in, i.e. (x,y)
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Solve:
Just add the equations,
:
4x^2 - 3y^2 = 24
2x^2 + 3y^2 = 30
------------------ adding eliminates y^2
6x^2 + 0 = 54;
x^2 = 54/6
x^2 = 9
x = Sqrt(9)
x = +/-3
:
Substitute 3 for x in one of the equations and find y:
;
Then check your solutions in the other equation