Our first step is to form a simple equation where \(x = 0.77..\). By multiplying both sides by \(10\) we can obtain another equation with \(10x = 7.77..\). Now we eliminate the recurring part of the decimal by subtracting \(x\) from \(10x\). \begin{aligned} x & = 0.77.. \\ 10x & = 7.77.. \\ 9x & = 7 \\ x & = {7 \over 9} \\ \end{aligned} So we have our answer \(0.77... = {7 \over 9} \).

The important part to remember is to get two equations in \(x\) where the recurring part after the decimal point is exactly the same.

Again our first step is write \(x = 0.277..\). Now multiplying by \(10\) gives us \(10x = 2.77..\). This time we need another equation to match the recurring part of the equation. So multiplying by \(10\) again gives \(100x = 27.77..\). Now we have two equations with the same recurring part we subtract one from the other as before. \begin{aligned} x & = 0.277.. \\ 10x & = 2.77.. \\ 100x & = 27.77.. \\ 90x & = 25 \\ x & = {25 \over 90} \\ x & = {5 \over 18} \\ \end{aligned} So in its lowest terms \(0.277.. = {5 \over 18} \).

As always our first step it to write \(x = 0.0855..\). This time to get two equations with the same recurring part after the decimal point we need to multiply by \(100\) and then \(1000\). This gives us: \begin{aligned} 100x & = 8.55.. \\ 1000x & = 85.55.. \\ 900x & = 77 \\ x & = {77 \over 900} \\ \end{aligned} This cannot be simplified any further so \(0.855.. = {77 \over 900}\).

This time the '\(234\)' part is the important bit. Now to get two multiples of x with 234 as the recurring part we need to multiply first by \(10\) and then another \(1000\) to move the digits 3 places to the right. \begin{aligned} x = 0.0234234... \\ 10x = 0.234234.. \\ 10000x = 234.234 \\ 9990x = 234 x \\ x = {234 \over 9990} \\ x = {13 \over 555} \\ \end{aligned} So \(0.0234 = {13 \over 555}\) as a fraction in its lowest terms.