What is the fractional equivalent of the repeating decimal #0.bar8#?

#0.bar8#

1 Answer
Jan 29, 2018

See a solution process below:

Explanation:

First, we can write:

#x = 0.bar8#

Next, we can multiply each side by #10# giving:

#10x = 8.bar8#

Then we can subtract each side of the first equation from each side of the second equation giving:

#10x - x = 8.bar8 - 0.bar8#

We can now solve for #x# as follows:

#10x - 1x = (8 + 0.bar8) - 0.bar8#

#(10 - 1)x = 8 + 0.bar8 - 0.bar8#

#9x = 8 + (0.bar8 - 0.bar8)#

#9x = 8 + 0#

#9x = 8#

#(9x)/color(red)(9) = 8/color(red)(9)#

#(color(red)(cancel(color(black)(9)))x)/cancel(color(red)(9)) = 8/9#

#x = 8/9#