Question

$2.3$ ($3$ repeating) as a fraction?

Answer 1

First, we can write:
$x=2.\overline{3}$
Next, we can multiply each side by $10$ giving:
$10x=23.\overline{3}$
Then we can subtract each side of the first equation from each side of the second equation giving:
$10x-x=23.\overline{3}-2.\overline{3}$
We can now solve for $x$ as follows:
$10x-1x=(23+0.\overline{3})-(2+0.\overline{3})$
$(10-1)x=23+0.\overline{3}-2-0.\overline{3}$
$9x=(23-2)+(0.\overline{3}-0.\overline{3})$
$9x=21+0$
$9x=21$
$\frac{9x}{9}=\frac{21}{9}$
$x=\frac{3\times 7}{3\times 3}$
$x=\frac{7}{3}$