6 3 x + 2 y + 1 2 x + 2 y + 7 2 x + 6 y + 4 3 2 x + 1 8 y + … = 6
6 4 x + 3 y + 1 8 8 x + 3 y + 5 4 1 6 x + 3 y + 1 6 2 3 2 x + 3 y + … = 8 1 0 7
Find the value of x + y .
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The first equation can be reduced to 2 x + 1 2 x + 7 2 x + 4 3 2 x + ...+ 3 y + 6 y + 1 2 y + 2 4 y +...= 6 and the second, 3 2 x + 9 4 x + 2 7 8 x + 8 1 1 6 x + . . . + 2 y + 6 y + 1 8 y + 5 4 y + . . . = 8 1 0 7 .
Now, we can observe a common ratio in all four sequences so using sum to infinity, we have 1 − 6 1 2 x + 1 − 2 1 3 y = 6 and 1 − 3 2 3 2 x + 1 − 3 1 2 y = 8 1 0 7 .
Now, 6 5 2 x + 2 1 3 y = 6 and 3 1 3 2 x + 3 2 2 y = 8 1 0 7
Or 5 3 x + 3 2 y = 6 and 2 x + 4 3 y = 8 1 0 7 . Solving these two equations you should get x = 5 and y = 2 9 .