First, expand the terms within parenthesis on the left side of the equation by multiplying each term within the parenthesis by #color(red)(4)# which is the term outside the parenthesis:
#6 + (color(red)(4) xx r) - (color(red)(4) xx 2) = r + 7#
#6 + 4r - 8 = r + 7#
#6 - 8 + 4r = r + 7#
#-2 + 4r = r + 7#
Next, add #color(red)(2)# and subtract #color(blue)(r)# from each side of the equation to isolate the #r# term while keeping the equation balanced:
#-2 + 4r + color(red)(2) - color(blue)(r) = r + 7 + color(red)(2) - color(blue)(r)#
#-2 + color(red)(2) + 4r - color(blue)(r) = r - color(blue)(r) + 7 + color(red)(2)#
#0 + 3r = 0 + 9#
#3r = 9#
Now, divide each side of the equation by #color(red)(3)# to solve for #r# while keeping the equation balanced:
#(3r)/color(red)(3) = 9/color(red)(3)#
#(color(red)(cancel(color(black)(3)))r)/cancel(color(red)(3)) = 3#
#r = 3#
