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Algebra Difficulty 4.4 AIME Find the answer
Suppose that x,y,z are real numbers such that x=y+z+2, y=z+x+1, and z=x+y+4. Compute x+y+z.
A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.
Solution
Adding all three equations gives x+y+z=2(x+y+z)+7 from which we find that x+y+z=−7.
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