How do you solve the system #y^2+x=2# and #3x+y=8#?

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Answer 1 · 2016-07-04T01:04:06

#x = 2 - y^2#

#:.3(2 - y^2) + y = 8#

#6 - 3y^2 + y = 8#

#0 = 3y^2 - y + 2#

#y = (-(-1) +- sqrt(-1^2 - 4 xx 3 xx 2))/(2 xx 3)#

#y = (1 +- sqrt(-23))/6#

#y = O/#

#:.# This systems of equations has no real solution.

Hopefully this helps!