First, add #color(red)(x)# and #color(blue)(19)# to each side of the equation to isolate the #x# term while keeping the equation balanced:
#-x - 11 + color(red)(x) + color(blue)(19) = x - 19 + color(red)(x) + color(blue)(19)#
#-x + color(red)(x) - 11 + color(blue)(19) = x + color(red)(x) - 19 + color(blue)(19)#
#0 + 8 = 2x - 0#
#8 = 2x#
Now, divide each side of the equation by #color(red)(2)# to solve for #x# while keeping the equation balanced:
#8/color(red)(2) = (2x)/color(red)(2)#
#4 = (color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2))#
#4 = x#
#x = 4#