Solve the system of equations algebraically

2x + 2y = 10
x + y = 5
x = 5 - y
Substitute for x in the other equation:-

(5 - y - 3)^2 + (y + 2)^2 = 16
(2 - y)^2 + (y + 2)^2 = 16
4 - 4y + y^2 + y^2 + 4y + 4 = 16
2y^2  + 8 = 16

2y^2 - 8 = 0
y^2 =  4

y = +/- 2

consider 2x  + 2y = 10

when y = -2,    2x  =  10 +4  giving    x = 7
when y  = 2,  2x  =  10 - 4  giving x =  3

so solution is  x = 7, y = -2  and x = 3,  y=2.