# How to make f the subject of 1/u + 1/v = 1/f and other fomulas that contain algebraic fractions.

Here you will learn how to rearrange a difficult formula which involves algebraic fractions.

**Example 1**

Make f the subject of 1/u + 1/v = 1/f

To do this first you will need to add together the two fractions on the left hand side. You can do this by making two equivalent fractions with a common denominator of uv.

Do this by multiplying the numerator and denominator of 1/u by v to give v/uv.

Also multiply the numerator and denominator of 1/v by u to give u/uv.

So 1/u + 1/v = 1/f gives:

v/uv + u/uv = 1/f

You can now add these two fractions together:

(u+v)/uv = 1/f

Both sides of the fraction can now be divided into 1 (so the fractions just flip over):

uv/(u+v) = f/1

So f = uv/(u+v)

**Example 2**

Make c the subject of 2/a - 3/b = 5/c

Like example 1, first you will need to subtract the two fractions on the left hand side. You can do this by making two equivalent fractions with a common denominator of ab.

Do this by multiplying the numerator and denominator of 2/a by b to give 2b/ab.

Also multiply the numerator and denominator of 3/b by a to give 3a/ab.

So 2/a - 3/b = 5/c gives:

2b/ab - 3a/ab = 5/c

You can now add these two fractions together:

(2b-3a)/ab = 5/c

Both sides of the fraction can now be divided into 1 (so the fractions just flip over):

ab/(2b-3a) = c/5

All you need to do now is multiply both sides of the equation by 5:

5ab/(2b-3a) = c

So the formula for c is c = 5ab/(2b-3a)

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## Comments 2 comments

Hi catman shouldn't this be C = 5ab/2b-3a

2b - 3a/ab = 5/c

c(2b-3a) = 5ab

c= 5ab/2b-3a