I am assuming the #1# and the #5# repeat as #0.151515...#
If it is just the #5# repeating you can use this same process.
First, we can write:
#x = 0.bar15#
Next, we can multiply each side by #100# giving:
#100x = 15.bar15#
Then we can subtract each side of the first equation from each side of the second equation giving:
#100x - x = 15.bar15 - 0.bar15#
We can now solve for #x# as follows:
#100x - 1x = (15 + 0.bar15) - 0.bar15#
#(100 - 1)x = 15 + 0.bar15 - 0.bar15#
#99x = 15 + (0.bar15 - 0.bar15)#
#99x = 15 + 0#
#99x = 15#
#(99x)/color(red)(99) = 15/color(red)(99)#
#(color(red)(cancel(color(black)(99)))x)/cancel(color(red)(99)) = (3 xx 5)/color(red)(3 xx 33)#
#x = (color(red)(cancel(color(black)(3))) xx 5)/color(red)(color(black)(cancel(color(red)(3))) xx 33)#
#x = 5/33#