math
posted by Jill .
Solve the system
x^2+y^24x+2y=20
4x +3y =5

The first one is a circle centred at (2,1), or
(x2)^2+(y+1)^2=20+4+1=25
or
C: (x2)^2+(y+1)^2=5^2
The second equation is a straight line.
L: 4x+3y=5
To solve the system, set
y=(54x)/3
and substitute into equation C
or
(x2)^2+((54x)/3+1)^2=25
Solve for x to get x=5 or x=1
Substitute into y=(54x)/3
to get
y=5, y=3
So the intersections are
(5,5), or (1,3)