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use elimination method to solve system of equations; 5u+3v=-15 2u+v=-6
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5u + 3v = -15 (1) 2u + 1v = - 6 (2) 6u + 3v = -18 (2)*3 1u + 0v = -3 (2)*3 - (1) Then back substitute into (1) 5*-3 + 3v = -15 -15 + 3v = -15 3v = 0 v = 0 Confirm using (2) 2*-3 + 1v = -6 -6 + 1v = -6 1v = 0 v = 1 u = -1, v = 0
Sorry u = -3, v = 0
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