First, isolate the #A# terms on the left side of the equation and the constants on the right side of the equation while keeping the equation balanced:
#10A + color(red)(2A) = 60 - 2A + color(red)(2A)#
#(10 + 2)A = 60 - 0#
#12A = 60#
Now we can solve for #A# by dividing each side of the equation by #color(red)(12)# which will also keep the equation balanced:
#(12A)/color(red)(12) = 60/color(red)(12)#
#(color(red)(cancel(color(black)(12)))A)/cancel(color(red)(12)) = 5#
#A = 5#