WAEC Mathematics 2018 Paper

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$\frac{1}{2}$