I'm assuming this is supposed to be .. x^2+y^2=5 2y+x=3 ..if so......
Using the second equation we can rearrange it to x = 3-2y ... and substituting this into the first equation, we have
(3 - 2y)^2 + y^2 = 5 simplify
4y^2 -12y + 9 + y^2 = 5 subtract 5 from both sides and simplify
5y^2 -12y +4 = 0
(5y -2) (y -2) = 0 and setting each factor to 0, we have that y =2/5 and y =2
And using x = 3-2y we have that x = 3-2(2/5) = 11/5 and x = 3-2(2) = -1
So, our solutions are (11/5, 2/5) and (-1, 2)
BTW......these are the intersection points of a line [2x + y =3] and a circle [x^2 + y^2 = 5] centered at the origin with a radius of √5
