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\(\frac{1}{2}\)