Recurring Proof Note $5$

I'll prove that

$y = 0.x_1\overline{x_2} =$ $\frac{x_1x_2 - x_1}{90}$

Let:

$y = 0.x_1\overline{x_2}$

$10y = x_1.\overline{x_2}$

$10y - y = 9y$

$x_1.\overline{x_2} - 0.x_1\overline{x_2} = x_1.x_2 - x_1$

$9y = x_1.x_2 - x_1$

$\frac{9y}{9}$ $=$ $\frac{x_1.x_2 - x_1}{9}$

$\frac{90y}{90}$ $=$ $\frac{x_1x_2 - x_1}{90}$

$y =$ $\frac{x_1x_2 - x_1}{90}$

Therefore, $y = 0.x_1\overline{x_2} =$ $\frac{x_1x_2 - x_1}{90}$

#Algebra

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Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Recurring Proof Note \(5\)

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