Posted by **amber** on Wednesday, May 4, 2011 at 10:48pm.

how to solve

y= (x-2)^2+4

4x+2y=14

algebraically

- geometry -
**MathMate**, Thursday, May 5, 2011 at 9:16am
By substitution:

y= (x-2)^2+4 ....(1)

4x+2y=14 ....(2)

Solve for y from (2) divided by 2:

2x+y=7

y=7-2x ...(2a)

Substitute (2a) in (1)

7-2x = (x-2)^2+4

Expand

7-2x = x^2-4x+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:

(1-2)^2+4 = 5 OK.

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