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Algebra Difficulty 2.7 Junior Find the answer

Suppose that xx and yy are real numbers that satisfy the two equations: x2+3xy+y2=909x^{2} + 3xy + y^{2} = 909 and 3x2+xy+3y2=12873x^{2} + xy + 3y^{2} = 1287. What is a possible value for x+yx+y?

A number or a short expression. Spacing and $ signs are ignored.

Solution

Since x2+3xy+y2=909x^{2} + 3xy + y^{2} = 909 and 3x2+xy+3y2=12873x^{2} + xy + 3y^{2} = 1287, then adding these gives 4x2+4xy+4y2=21964x^{2} + 4xy + 4y^{2} = 2196. Dividing by 4 gives x2+xy+y2=549x^{2} + xy + y^{2} = 549. Subtracting this from the first equation gives 2xy=3602xy = 360, so xy=180xy = 180. Substituting xy=180xy = 180 into the first equation gives x2+2xy+y2=729x^{2} + 2xy + y^{2} = 729, which is (x+y)2=272(x+y)^{2} = 27^{2}. Therefore, x+y=27x+y = 27 or x+y=27x+y = -27. This also shows that x+yx+y cannot equal any of 39, 29, 92, and 41.

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