First, multiple each side of the equation by #color(red)(12)# to eliminate the fractions while keeping the equation balanced:
#color(red)(12)(5/6x + 1/2) = color(red)(12) xx 1/4(x - 4)#
#(color(red)(12) xx 5/6x) + (color(red)(12) xx 1/2) = cancel(color(red)(12)) 3 xx 1/color(red)(cancel(color(black)(4)))(x - 4)#
#(cancel(color(red)(12)) 2 xx 5/color(red)(cancel(color(black)(6)))x) + (cancel(color(red)(12)) 6 xx 1/color(red)(cancel(color(black)(2)))) = 3(x - 4)#
#10x + 6 = 3(x - 4)#
Then, expand the terms within parenthesis on the right side of the equation:
#10x + 6 = (3 xx x) - (3 xx 4)#
#10x + 6 = 3x - 12#
Next, subtract #color(red)(3x)# and #color(blue)(6)# from each side of the equation to isolate the #x# term while keeping the equation balanced:
#10x + 6 - color(red)(3x) - color(blue)(6) = 3x - 12 - color(red)(3x) - color(blue)(6)#
#10x - color(red)(3x) + 6 - color(blue)(6) = 3x - color(red)(3x) - 12 - color(blue)(6)#
#7x + 0 = 0 - 18#
#7x = -18#
Now, divide each side of the equation by #color(red)(7)# to solve for #x# while keeping the equation balanced:
#(7x)/color(red)(7) = -18/color(red)(7)#
#(color(red)(cancel(color(black)(7)))x)/cancel(color(red)(7)) = -18/7#
#x = -18/7#