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2015 AMC 10A Problem 16

Problem 16 of 25IntermediateAlgebra

If y+4=(x−2)2,y+4 = (x-2)^2, x+4=(y−2)2, x+4 = (y-2)^2, and x≠y,x \neq y, what is the value of x2+y2x^2+y^2

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Solution

Adding the two equations gives us x2+y2−4x−4y+8 x^2 + y^2 - 4x - 4y + 8 =x+y+8. = x + y + 8. We can rearrange this equation to get x2+y2=5(x+y). x^2 + y^2 = 5(x + y). We can then subtract them to get x2−y2−4x+4y=y−x. x^2 - y^2 - 4x + 4y = y - x. Once again rearranging, we can find x2−y2=3(x−y). x^2 - y^2 = 3(x - y). We have that x≠y,x \neq y, which means that we can divide both sides by x−y.x - y. This gives us x+y=3 x + y = 3 x2+y2=15. x^2 + y^2 = 15. Thus, B is the correct answer.
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Tagged: system of equations · symmetry (algebra) · algebraic manipulation

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