Describe the error in solving the system of equations. = 3 + 2r 2 %3D2
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= 3 + 2r 2х — у %3D2 x? + 2x – (-2x + 2) = 3 x + 4x – 2 = 3 x + 4x – 5 = 0 (x + 5)(x – 1) = 0 x = -5, 1 When x = -5, y = -2(-5) + 2 = -8, and when x = 1, y = -2(1) + 2 = 0. So, the solutions are (-5, –8) and (1, 0).
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