Simultaneous Equations (Reducible to Quadratics)

Simultaneous Equations (Reducible to Quadratics)?
We were doing equations reducible to quadratics and i got this question, i got this and have no idea what to do.
I DO NOT UNDERSTAND WHY THE ANSWER IS (1,3)

x^2+y^2=10
x+2y=7

Solve by substitution.

x + 2y = 7
x = 7 - 2y

x^2 + y^2 = 10 (substitute 7 - 2y for x)
(7 - 2y)^2 + y^2 = 10
4y^2 - 28y + 49 + y^2 = 10
5y^2 - 28y + 49 = 10
5y^2 - 28y + 39 = 0
(5y - 13)(y - 3) = 0
5y - 13 = 0 or y - 3 = 0
y = 13/5 or y = 3

When y = 13/5:
x = 7 - 2y
x = 7 - 2(13/5)
x = 7 - 26/5
x = 35/5 - 26/5
x = 9/5

When y = 3:
x = 7 - 2y
x = 7 - 2(3)
x = 7 - 6
x = 1

The solutions (x, y) are (9/5, 13/5) and (1, 3). (1, 3) is not the only answer.

substitute x=7-2y in first equation
u will get
49+5y^2-28y=10
or 5y^2-28y+39=0
solve for y
u will get y=3 and y=13/5
now substitute y=3
in both equations
from 1st u will get 1 and-1
and from 2nd u will get 1
common in both is 1so (1,3) is a solution

now substitute y=13/5 in both

from both u will get x=9/5 a solution
so (9/5,13/5) is also a solution