SOLUTION: Solve the system algebraically. x^2+y^2+2x+4y-8=0 3x+2y=6 I wasn't really sure how to solve this so I started, but I sort of got stuck... 2y=-3x+6 y=-(3/2)x+3 x^2+(-3/2x+

Question 695812: Solve the system algebraically.
x^2+y^2+2x+4y-8=0
3x+2y=6
I wasn't really sure how to solve this so I started, but I sort of got stuck...
2y=-3x+6
y=-(3/2)x+3
x^2+(-3/2x+3)^2+2x+4(-3/2x+3)-8=0
...then I got stuck.
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
You're not that far from the solution:

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