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!

How do you solve the system y^2+x=2y^2+x=2 and 3x+y=83x+y=8?