Prove algebraically that the recurring decimal 0.6˙8˙1˙0.\dot{6}\dot{8}\dot{1}0.6˙8˙1˙ can be written as 227333\displaystyle \frac{227}{333}333227.
Hence, find 1−0.6˙8˙1˙1-0.\dot{6}\dot{8}\dot{1}1−0.6˙8˙1˙
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