Use back substitution to solve each of the following systems of equations: (a) x1 3x2 = 2 2x2 = 6 (b) x1 + x2 + x3 = 8 2x2 + x3 = 5 3x3 = 9 (c) x1 + 2x2 + 2x3 + x4 = 5 3x2 + x3 2x4 = 1 x3 + 2x4 = 1 4x4 = 4 (d) x1 + x2 + x3 + x4 + x5 = 5 2x2 + x3 2x4 + x5 = 1 4x3 + x4 2x5 = 1 x4 3x5 = 0 2x5 = 2

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