Question

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

Answer 1

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)`

Answer 2

`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)`