x^2+y^2=5 (circle)
3x+y=7 (line)
Rearranging the second equation, we have y = 7-3x. Now, we can substitute for y in the first one....so we have
x^2 + (7 - 3x)^2 = 5 simplifying gives us
x^2 + 49 - 42x + 9x^2 = 5 subtract 5 from both sides
x^2 + 49 - 42x + 9x^2 - 5 = 0 simplify the left side, we have
10x^2 - 42x + 44 = 0 divide by 2 on both sides
5x^2 - 21x + 22 = 0 factor
(5x -11) (x - 2) = 0 set each factor to 0
5x - 11 = 0 x - 2 = 0
x = 11/5 x = 2
Now we can use y = 7-3x to find y
So we have
y = 7 - 3(11/5) and y = 7 - 3(2)
y = 7 - 33/5 and y = 7 - 6
y = 35/5 - 33/5 and y = 1
y = 2/5 and y = 1
So our solutions are (11/5 , 2/5) and (2 , 1)
Note that this is just the intersection of the line and the circle in two points.....
