SOLUTION: Solve the systems of equations algebraically, test each solution for validity, and then graph the system to check your number of solutions: x^2-2y^2=2 xy=2

Algebra ->  Trigonometry-basics -> SOLUTION: Solve the systems of equations algebraically, test each solution for validity, and then graph the system to check your number of solutions: x^2-2y^2=2 xy=2      Log On


   

Question 1161845: Solve the systems of equations algebraically, test each solution for validity, and then graph the system to check your number of solutions:
x^2-2y^2=2
xy=2
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
Solve the systems of equations algebraically, test each solution for validity, and then graph the system to check your number of solutions:
x%5E2-2y%5E2=2...eq.1
xy=2...eq.2
------------------------------
xy=2...eq.2........solve for x
x=2%2Fy+..eq.2a

go to
x%5E2-2y%5E2=2...eq.1, substitute+x
%282%2Fy+%29%5E2-2y%5E2=2... solve for+y
4%2Fy%5E2-2y%5E2=2....both sides multiply by y%5E2
4-2y%5E4=2y%5E2
0=2y%5E4%2B2y%5E2-4......both sides divide by 2
0=y%5E4%2By%5E2-2....factor
0=y%5E4-y%5E2%2B2y%5E2-2
%28y%5E4-y%5E2%29%2B%282y%5E2-2%29=0
y%5E2%28y%5E2-1%29%2B2%28y%5E2-1%29=0
%28y%5E2%2B2%29%28y%5E2-1%29=0
%28y+-+1%29+%28y+%2B+1%29+%28y%5E2+%2B+2%29=0
solutions:
if %28y+-+1%29+=0 ->y=1
if %28y+%2B+1%29+=0+->y=-1
if %28y%5E2+%2B+2%29=0+->y%5E2=-2-> y= ± sqrt%28-2%29= ± sqrt%282%29%2Ai
go to x=2%2Fy...eq.2a , substitute y=1
x=2%2F1
x=2
or
x=2%2Fy+, substitute y=-1
x=2%2F-1
x=-2


real solutions are:
x=2, y=1
and
x=-2, y=-1

test each solution for validity:
x%5E2-2y%5E2=2.......x=2, y=1
2%5E2-2%2A1%5E2=2
4-2=2
2=2-> solution valid

xy=2........x=-2, y=-1
-2%2A%28-1%29=2
2=2-> solution valid

graph: