Given:#" "color(magenta)(2)color(blue)((3+m))-1=3m+11#
Multiply everything inside the bracket by the 2 that is outside giving:
#(color(magenta)(2xx) color(blue)(3))color(blue)(+)(color(magenta)(2xx)color(blue)(m))-1" "=" "3m+11#
#" "6" "+" "2m" "-1" "=" "3m+11#
#" "5+2m" "=" "3m+11#
Subtract #color(red)(2m)# from both sides (removes it from the left)
#" "color(green)(5+2mcolor(red)(-2m)" "=" "3mcolor(red)(-2m)+11)#
#" "5+" "0" "=" "m" "+11#
#" "5" "=" "m+11#
Subtract #color(red)(11)# from both sides (removes it from the right)
#" "color(green)(5color(red)(-11)" "=" "m+11color(red)(-11))#
#" "-6" "=" "m+0#
To comply with convention write this so that the unknown variable (#m#) is on the left
#" "m" "=" "-6#
