How do you solve #3x^2 - y^2 =1 1# and #x^2 + 2y = 2# using substitution?

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Answer 1 · 2016-05-03T01:20:25

#x^2 = 2 - 2y#

#x = sqrt(2 - 2y)#

#3(sqrt(2 - 2y))^2 - y^2 = 11#

#3(2 - 2y) - y^2 = 11#

#6 - 6y - y^2 = 11#

#0 = y^2 + 6y + 5#

#0 = (y + 1)(y + 5)#

#y = -1 and -5#

#x^2 + 2(-1) = 2 or x^2 + 2(-5) = 2#

#x^2 = 4 or x^2 = 12#

#x = 2 or x = 2sqrt(3)#

Your solution sets are #{2, -1} and {2sqrt(3), -5}#.

Hopefully this helps!