Because both sides of the equation and single fractions we can "flip" the fractions giving:
#(r + 4)/2 = r/4#
Next, multiply each side of the equation by #color(red)(4)# to eliminate the fractions while keeping the equation balanced:
#color(red)(4) xx (r + 4)/2 = color(red)(4) xx r/4#
#cancel(color(red)(4))2 xx (r + 4)/color(red)(cancel(color(black)(2))) = cancel(color(red)(4)) xx r/color(red)(cancel(color(black)(4)))#
#2(r + 4) = r#
Then, expand the term on the left side of the equation by multiplying each term within the parenthesis by the term outside the parenthesis:
#(2 xx r) + (2 xx 4) = r#
#2r + 8 = r#
Now, subtract #color(red)(8)# and #color(blue)(r)# from each side of the equation to solve for #r# while keeping the equation balanced:
#-color(blue)(r) + 2r + 8 - color(red)(8) = -color(blue)(r) + r - color(red)(8)#
#-1color(blue)(r) + 2r + 0 = 0 - color(red)(8)#
#(-1 + 2)r = -8#
#1r = -8#
#r = -8#