geometry
posted by amber .
how to solve
y= (x2)^2+4
4x+2y=14
algebraically

By substitution:
y= (x2)^2+4 ....(1)
4x+2y=14 ....(2)
Solve for y from (2) divided by 2:
2x+y=7
y=72x ...(2a)
Substitute (2a) in (1)
72x = (x2)^2+4
Expand
72x = x^24x+4+4
x^2 2x +1 = 0 ....(1a)
Solve 1(a) to get:
x=1 (duplicity 2)
Now substitute x=1 in (2a) to get
y=5
Substitute x=1,y=5 in (1) to check:
(12)^2+4 = 5 OK.