Word Count : 248

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In ℝ³, for the equation x − π = 5, solving by substitution gives x = 5 + π. Since there are no restrictions on y and z, the solution set is {(5 + π, y, z) : y, z ∈ ℝ}. For the system 2x − y + z = 1, x − 2y + z = 3, y = √2 __________ ____ _____ ______ ______ _________ ________.
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