Problem no. 23

The first equation can be reduced to $\frac{x}{2}$ + $\frac{x}{12}$ + $\frac{x}{72}$ + $\frac{x}{432}$ + ...+ $\frac{y}{3}$ + $\frac{y}{6}$ + $\frac{y}{12}$ + $\frac{y}{24}$ +...= 6 and the second, $\frac{2x}{3} + \frac{4x}{9} + \frac{8x}{27} + \frac{16x}{81} +... + \frac{y}{2} + \frac{y}{6} + \frac{y}{18} + \frac{y}{54} + ... = \frac{107}{8}$ .

Now, we can observe a common ratio in all four sequences so using sum to infinity, we have $\frac{\frac{x}{2}}{1-\frac{1}{6}} + \frac{\frac{y}{3}}{1-\frac{1}{2}} = 6$ and $\frac{\frac{2x}{3}}{1-\frac{2}{3}} + \frac{\frac{y}{2}}{1-\frac{1}{3}} = \frac{107}{8}$ .

Now, $\frac{\frac{x}{2}}{\frac{5}{6}} + \frac{\frac{y}{3}}{\frac{1}{2}} = 6$ and $\frac{\frac{2x}{3}}{\frac{1}{3}} + \frac{\frac{y}{2}}{\frac{2}{3}} = \frac{107}{8}$

Or $\frac{3x}{5} + \frac{2y}{3} = 6$ and $2x + \frac{3y}{4} = \frac{107}{8}$ . Solving these two equations you should get x = 5 and y = $\frac{9}{2}$ .