First, isolate the #a# terms on one side of the equation and the constants on the other side of the equation while keeping the equation balanced:
#3a + 13 - color(red)(3a + 8) = 9a - 8 - color(red)(3a + 8)#
#3a - 3a + 13 + 8 = 9a - 3a - 8 + 8#
#0 + 13 + 8 = 9a - 3a - 0#
#13 + 8 = 9a - 3a#
Next, we can combine like terms:
#21 = (9 - 3)a#
#21 = 6a#
Now, solve for #a# while keeping the equation balanced:
#21/color(red)(6) = (6a)/color(red)(6)#
#(3 xx 7)/(3 xx 2) = (color(red)(cancel(color(black)(6)))a)/color(red)(cancel(color(black)(6)))#
#3/3 xx 7/2 = a#
#1 xx 7/2 = a#
#7/2 = a#