How do you solve the system #6x^2+3y^2=12# and #y=-x+2#?

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Answer 1 · 2016-08-01T19:53:56

The Soln. Set #={(0,2), (4/3,2/3)}#

We subst. #y# from the Second eqn. into the First one, to get,

#6x^2+3(2-x)^2=12#

#:. 6x^2+3(4-4x+x^2)=12#

#:. 9x^2-12x=12-12=0#.

#:. 3x(3x-4)=0#

#:. x=0, or, x=4/3#

accordingly, #y=2, or, 2/3#

#:.# The Soln. Set #={(0,2), (4/3,2/3)}#