# geometry

how to solve
y= (x-2)^2+4
4x+2y=14
algebraically

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asked by amber
1. 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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posted by MathMate

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