How do you solve for a in #1/a + 1/b = 1/f #?

2 Answers
Aug 7, 2017

Answer:

See a solution process below:

Explanation:

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)#

Aug 7, 2017

Answer:

#color(magenta)(a=(bf)/(b-f)#

Explanation:

#1/a+1/b=1/f#

#:.(bf+af=ab)/(abf)#

multiply both sides by #abf#

#:.bf+af=ab#

#:.ab-af=bf#

#:.a(b-f)=bf#

#:.color(magenta)(a=(bf)/(b-f)#