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Question 699382: solve the following system
x^2+y=3
xy=2
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! x^2+y=3
xy=2
y=2/x
substitute y in first equation
x^2+y=3
x^2+(2/x) =3
multiply equation by x
x^3+2=3x
x^3-3x+2=0
to find the zeros of the equation
plug x=1
x=1 is one solution
(x-1)(x^2+x-2)=0
(x-1)(x^2-2x+x-2))=0
(x-1)(x(x-2)+1(x-2)) =0
(x-1)(x+1)(x-2) =0
x=1,-1,2
plug the three vaues of x to get corresponding values of y
you get three solutions
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