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

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Answer 1 · 2018-03-07T19:36:01

Pairs of the solution are #(-2sqrt3, -5)#, #(-2, -1)#, #(2, -1)# and #(2sqrt3, -5)#

#3*(x^2+2y)-(3x^2-y^2)=2*3-11#

#3x^2+6y-3x^2+y^2=-5#

#y^2+6y=-5#

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

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

#y_1=-5# and #y_2=-1#

For #y=-5#, #x^2+2*(-5)=2#, so #x^2=12#

Hence #x_1=-2sqrt3# and #x_2=2sqrt3#

For #y=-1#, #x^2+2*(-1)=2#, so #x^2=4#

Hence #x_3=-2# and #x_4=2#

Thus, pairs of the solution are #(-2sqrt3, -5)#, #(-2, -1)#, #(2, -1)# and #(2sqrt3, -5)#