i need to solve a simultaneous equation with one linear and one quadratic equation?

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.....