First, multiply each side of the equation by #color(red)(10)# to eliminate the fractions while keeping the equation balanced. #color(red)(10)# was selected because it is the LCM (Least Common Multiple) of the two fractions:
#color(red)(10) xx (x + 4)/2 = color(red)(10)(4 - (x + 1)/5)#
#cancel(color(red)(10))5 xx (x + 4)/color(red)(cancel(color(black)(2))) = (color(red)(10) xx 4) - (color(red)(10) xx (x + 1)/5)#
#5(x + 4) = 40 - (cancel(color(red)(10))2 xx (x + 1)/color(red)(cancel(color(black)(5))))#
#(5 xx x) + (5 xx 4) = 40 - (2(x + 1))#
#5x + 20 = 40 - (2 xx x) - (2 xx 1)#
#5x + 20 = 40 - 2x - 2#
#5x + 20 = 40 - 2 - 2x#
#5x + 20 = 38 - 2x#
Next, subtract #color(red)(20)# and add #color(blue)(2x)# to each side of the equation to isolate the #x# term while keeping the equation balanced:
#5x + color(blue)(2x) + 20 - color(red)(20) = 38 - color(red)(20) - 2x + color(blue)(2x)#
#(5 + color(blue)(2))x + 0 = 18 - 0#
#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#
