How we can do this is by making the number #(0.7bar8)# equivalent to a pro numeral, and for this example, we'll use #x#

Note: #barx# just means that #x# is a reccurring /repeating number, so #0.7bar8 = 0.788888888888888888888...#.

So now we have:

#x = 0.7888.... = 0.7bar8#

What we can do is multiply #x# by #100# to get #100x#, and obviously we have to do that to the other side.

#x xx 100 = 78.bar8 xx 100#

#100x = 78.bar8#

#10x = 7.bar8#

The reason why we do this is because now have two numbers, #color(brown)(100x = 78.bar8#, and #color(brown)(10x = 7.bar8#, so now we can cancel out the two repeating decimals, and then subtract the second number from the first, to get a whole integer.

#(100x = 78. cancel(88bar8)) - (10x = 7. cancel(88bar8)) = (90x = 71)#

Now we can find #x# by using algebra.

#90x = 71#

Divide each side by #90# to find #x#

#(color(red)(cancel(color(black)90))x)/cancel(color(red)(90)) = 71/90#

#x= 71/90#

#color(blue)(0.7bar8 = 71/90#

Because we can not simplify any further, this is our final answer.

I got all of this information from Khan Academy's videos on this , you can check it out here:

Converting Repeating Decimals to Fractions 1

Converting Repeating Decimals to Fractions 2

Hope this Helps :)