Solve the simultaneous, linear and quadratic equation
X^2 + y^2 = 13 ............ (1)
X + y = -1 ..............(2)
We can solve this system of equations by substitution or elimination method.
Substituting equation (2) into equation (1), we get:
(X + y)^2 + y^2 = 13
Expanding the equation, we have:
X^2 + 2xy + y^2 + y^2 = 13
Rearranging, we have:
2Xy + 2y^2 = 13 - X^2
Dividing both sides by 2, we get:
Xy + y^2 = (13 - X^2)/2 ............(3)
Now we have two equations:
X + y = -1 ..............(2)
Xy + y^2 = (13 - X^2)/2 ............(3)
We can solve equation (2) for X:
X = -1 - y
Substituting this into equation (3), we get:
(-1 - y)y + y^2 = (13 - (-1 - y)^2)/2
-y - y^2 + y^2 = 13/2 - (1 + 2y + y^2)/2
-y = 13/2 - 1/2 - y
y - y = 13/2 - 1/2
0 = 12/2
0 = 6
This is a contradiction, so there is no solution to the system of equations.