If x : y : z = 3 : 3 : 4, evaluate \(\frac{9x + 3y}{6x - 2y}\) \(\frac{x}{y}\) = \(\frac{2}{3}\) and \(\frac{y}{z}\) = \(\frac{3}{4}\) Thus; x = \(\frac{2}{3}T_1\) and z = \(\frac{3}{5}T_1\) y = \(\frac{3}{7}T_2\) and z = \(\frac{4}{7}T_2\) Using y = y \(\frac{3}{5}T_1\) = \(\frac{3}{7}T_2\); \(\frac{T_1}{T_2}\) = \(\frac{3}{7}\) x \(\frac{5}{3}\) \(\frac{T_1}{T_2}\) = \(\frac{15}{21}\) \(T_1\) = 15 and \(T_2\) = 21 Therefore; x = \(\frac{2}{5}\) x 15 = 6 y = \(\frac{3}{5}\) x 15 = 9 y = \(\frac{3}{7}\) x 21 = 9 (again) z = \(\frac{4}{7}\) x 21 = 12 Hence; \(\frac{9x + 3y}{6z - 2y}\) = \(\frac{9(6) + 3(9)}{6(12) - 2(9)}\) \(\frac{54 + 27}{72 - 18}\) = \(\frac{81}{54}\) = \(\frac{3}{2}\) = 1\(\frac{1}{2}\)