First, subtract #color(red)(1/8)# and add #color(blue)(1/3x)# to each side of the equation to isolate the #x# term while keeping the equation balanced:
#color(blue)(1/3x) + 1/2x + 1/8 - color(red)(1/8) = color(blue)(1/3x) - 1/3x + 1/4 - color(red)(1/8)#
#color(blue)(1/3x) + 1/2x + 0 = 0 + 1/4 - color(red)(1/8)#
#color(blue)(1/3x) + 1/2x = 1/4 - color(red)(1/8)#
We must next put the fractions over common denominators so we can add or subtract them:
#(2/2 xx color(blue)(1/3x)) + (3/3 xx 1/2x) = (2/2 xx 1/4) - color(red)(1/8)#
#2/6x + 3/6x = 2/8 - color(red)(1/8)#
#(2 + 3)/6x = (2 - 1)/8#
#5/6x = 1/8#
Now, multiply each side of the equation by #color(red)(6)/color(blue)(5)# to solve for #x# while keeping the equation balanced:
#color(red)(6)/color(blue)(5) xx 5/6x = color(red)(6)/color(blue)(5) xx 1/8#
#cancel(color(red)(6))/cancel(color(blue)(5)) xx color(blue)(cancel(color(black)(5)))/color(red)(cancel(color(black)(6)))x = 6/40#
#x = (2 xx 3)/(2 xx 20)#
#x = (color(red)(cancel(color(black)(2))) xx 3)/(color(red)(cancel(color(black)(2))) xx 20)#
#x = 3/20#
