Find the equations of the tangent line to the curve. (Please show me the complete solution. Thanks.) Solution at x = 1, 2(1)2 + 3y2 - (1)y = 4 3y2 - y - 2 = 0 (3y + 2)(y - 1) = 0 y = -2/3 and 1 find slope: differentiate 2x2 + 3y2 - xy = 4 4x + 6yy\' - y - xy\' = 0 at x = 1, y = -2/3: 4 - 4y\' + 2/3 - y\' = 0, y\' = 14/15 equation of the tangent line: y = -2/3 + 14(x - 1)/15 = (14x - 24)/15 at x = 1, y = 1: 4 + 6y\' - 1 - y\' = 0, y\' = -3/5 equation of the tangent line: y = 1 - 3(x - 1)/5 = (8 - 3x)/5.