Welcome to the final stop in Mastery of Repeating Decimals! Over the first three lessons you built a powerful toolkit: reading bar notation, identifying repeating blocks in mixed expansions, and using the algebraic elimination method to convert pure repeating decimals into exact fractions. Now we are ready for the capstone skill — converting mixed repeating decimals into exact fractions.

A mixed repeating decimal has one or more non-repeating digits after the decimal point before the cycle begins, like $0.1\overline{6}$ or $0.91\overline{6}$. This lesson builds directly on the four-step technique from Lesson 3. The core idea stays the same — multiply, subtract, and cancel the repeating tail — but the new wrinkle is that mixed decimals require two different powers of ten instead of one. By the end of this lesson, you will handle any mixed repeating decimal with confidence.

In the previous lesson, converting a pure repeating decimal like $0.\overline{7}$ required just one multiplication by $10$. We multiplied, subtracted the original, and the repeating tails canceled neatly. That worked because the cycle started immediately after the decimal point, so a single shift by one block length was enough to line everything up.

Now consider $0.1\overline{6} = 0.16666\ldots$ The repeating block () does not begin until after the prefix digit . If we try the old approach and multiply only by , we get Subtracting the original does cancel the repeating part, giving , but the right side is a decimal rather than a whole number.

Let's see the two-shift method in full detail with $0.1\overline{6} = 0.16666\ldots$ Here the non-repeating prefix has 1 digit and the repeating block has 1 digit.

Step 1 — Define a variable.

Step 2 — Apply the larger shift. Multiply by $$ to move past the prefix one full block:

Let's try another example with the same structure — one prefix digit and a one-digit block — and move through it a bit more briskly.

Let $x = 0.8\overline{3} = 0.83333\ldots$ The prefix has $1$ digit and the block has $1$ digit, so we multiply by and by :

When the repeating block has more than one digit, the larger power of ten increases to match. Let's convert $0.1\overline{72} = 0.172727272\ldots$, which has a 1-digit prefix and a 2-digit repeating block.

• Larger shift: $10^{1+2} = 10^3 = 1{,}000$

What happens when the prefix is longer than one digit? The same logic applies — we simply increase the smaller power of ten. Let's convert $0.91\overline{6} = 0.91666\ldots$, which has a 2-digit prefix ($91$) and a 1-digit repeating block ($6$).

• Larger shift: ${10}^{}$

Let's collect everything into a tidy summary. For a mixed repeating decimal with $p$ non-repeating digits after the decimal point and $r$ repeating digits:

In this lesson, we extended the algebraic elimination method to mixed repeating decimals by introducing a second power of ten. The smaller shift moves past the non-repeating prefix, the larger shift moves past the prefix plus one full repeating block, and subtracting the two cancels the repeating tail cleanly. We practiced with prefixes and blocks of different lengths and distilled the approach into a general formula.

Now it is time to put this technique into action! In the upcoming practice exercises, you will complete a guided conversion step by step, write full derivations from scratch, tackle mixed decimals of increasing complexity, and even convert a real-world measurement into an exact fraction. Let's turn that fresh knowledge into lasting mastery!