First, subtract #color(red)(1/b)# from each side of the equation to isolate the #a# term while keeping the equation balanced:
#1/a + 1/b - color(red)(1/b) = 1/f - color(red)(1/b)#
#1/a + 0 = 1/f - 1/b#
#1/a = 1/f - 1/b#
Next, subtract the fractions on the right side of the equation after putting each fraction over a common denominator by multiplying each fraction by the appropriate form of #1#:
#1/a = (b/b xx 1/f) - (1/b f/f)#
#1/a = b/(bf) - f/(bf)#
#1/a = (b - f)/(bf)#
We can now "flip" the fraction on each side of the equation to solve for #a# while keeping the equation balanced:
#a/1 = (bf)/(b - f)#
#a = (bf)/(b - f)#
If you require the more rigorous process to solve for #a# see below:
Multiply each side of the equation by #abf# to eliminate the fractions while keeping the equation balanced:
#abf xx 1/a = abf xx (b - f)/(bf)#
#color(red)(cancel(color(black)(a)))bf xx 1/color(red)(cancel(color(black)(a))) = acolor(red)(cancel(color(black)(bf))) xx (b - f)/color(red)(cancel(color(black)(bf)))#
#bf = a(b - f)#
Now, divide each side of the equation by #color(red)(b - f)# to solve for #a# while keeping the equation balanced:
#(bf)/color(red)(b - f) = (a(b - f))/color(red)(b - f)#
#(bf)/(b - f) = (acolor(red)(cancel(color(black)((b - f)))))/cancel(color(red)(b - f))#
#(bf)/(b - f) = a#
#a = (bf)/(b - f)#