How do you solve the system #3x^2-y^2=-6# and #y=2x+1#?

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Answer 1 · 2016-08-29T17:34:59

#(x,y)=(1,3), or, (-5,-9)#.

We submit #y=2x+1# in the first eqn. and get,

#3x^2-(2x+1)^2+6=0#

#rArr -x^2-4x+5=0, or, x^2+4x-5=0#.

#rArr (x-1)(x+5)=0#

#rArr x=1, or, -5#.

#rArr y=2x+1=3, or, -9#

Therefore, the solns. are #(x,y)=(1,3), or, (-5,-9)#, which satisfy the set

of eqns.