SOLUTION: I would appreciate your help in solving this nonlinear equation: Problem: (x+2)^2+(y-2)^2=13 (1) ( 2x+y=6 (2) Thank you greatly for all your help.

Algebra ->  Linear-equations -> SOLUTION: I would appreciate your help in solving this nonlinear equation: Problem: (x+2)^2+(y-2)^2=13 (1) ( 2x+y=6 (2) Thank you greatly for all your help.      Log On


   

Question 111307: I would appreciate your help in solving this nonlinear equation:
Problem: (x+2)^2+(y-2)^2=13 (1)
( 2x+y=6 (2)

Thank you greatly for all your help.
Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
use the second equation to find y (y=6-2x) and then substitute this value into the first equation

(x+2)^2+(6-2x-2)^2=13 ___ x^2+4x+4+16-16x+4x^2=13 ___ 5x^2-12x+7=0 ___ (5x-7)(x-1)=0

plug the x values back into the second equation to find the corresponding y values ___ there will be two pairs

this is a line intersecting a circle (at two points)