First, we can multiply each side of the equation by #color(red)(12)color(blue)(x)# to eliminate the fractions while keeping the equation balanced. #color(red)(12)color(blue)(x)# is the Least Common Denominator for the three fractions:
#color(red)(12)color(blue)(x)((1/4) + (1/x)) = color(red)(12)color(blue)(x) xx (1/3)#
#(color(red)(12)color(blue)(x) xx (1/4)) + (color(red)(12)color(blue)(x) xx (1/x)) = cancel(color(red)(12))4color(blue)(x) xx (1/color(red)(cancel(color(black)(3))))#
#(cancel(color(red)(12))3color(blue)(x) xx (1/color(red)(cancel(color(black)(4))))) + (color(red)(12)cancel(color(blue)(x)) xx (1/color(blue)(cancel(color(black)(x))))) = 4x#
#3x + 12 = 4x#
Now, subtract #color(red)(3x)# from each side of the equation to solve for #x# while keeping the equation balanced:
#-color(red)(3x) + 3x + 12 = -color(red)(3x) + 4x#
#0 + 12 = (-color(red)(3) + 4)x#
#12 = 1x#
#12 = x#
#x = 12#