Rewrite the bases to express everything in terms of 2 and 3:\n\[ 2^{x + \frac{1}{2}} \times (2^2)^{y - \frac{5}{6}} = 3^{x - \frac{1}{2}} \times (3^2)^{y - \frac{1}{3}} \]\n\[ 2^{x + \frac{1}{2}} \times 2^{2y - \frac{5}{3}} = 3^{x - \frac{1}{2}} \times 3^{2y - \frac{2}{3}} \]\n\nCombine the exponents:\n\[ 2^{x + 2y + \frac{1}{2} - \frac{5}{3}} = 3^{x + 2y - \frac{1}{2} - \frac{2}{3}} \]\n\nSimplify the constant terms:\n\[ \frac{1}{2} - \frac{5}{3} = \frac{3 - 10}{6} = -\frac{7}{6} \]\n\[ -\frac{1}{2} - \frac{2}{3} = \frac{-3 - 4}{6} = -\frac{7}{6} \]\n\nSo the equation becomes:\n\[ 2^{x + 2y - \frac{7}{6}} = 3^{x + 2y - \frac{7}{6}} \]\n\nSince the bases (2 and 3) are different, the only way for their powers to be equal is if the exponent itself is zero:\n\[ x + 2y - \frac{7}{6} = 0 \]\n\nMultiply the entire equation by 6:\n\[ 6x + 12y - 7 = 0 \]