1. Use back substitution to solve each of the following systems of equations: - (a) x13x2 = 2 (b) x1 + x2 + x3 = 8 2x2 = 6 2x2 + x3 = 5 3x3 = 9 (c) x1+2x2 + 2x3 + x4 = 5 3x2 -x3 + 2x4 == -1 x32x4 = 1 4x4 = 4 (d) x1 + x2 + 2x2 x3 + x4 + x5 = 5 x3 2x4 + x5 = 1 4x3x42x5 = 1 x4 - 3x5 = 0 2x5 = 2