# Complex Fraction

How to solve a complex fraction: formula, 2 examples, and their solutions.

## Formula

A complex fraction is a fraction

whose numerator or denominator is also a fraction

(or both are also fractions).

To solve a complex fraction [a/b]/[c/d],

write the product of the outer factors,

ad,

in the numerator

and write the product of the inner factors,

bc,

in the denominator:

[a/b]/[c/d] = ad/bc.

## Example[(x + 2)/x] / [6/(x - 1)]

Write the product of the outer factors,

(x + 2)(x - 1),

in the numerator.

And write the product of the inner factors,

x⋅6 = 6x,

in the denominator.

So

(x + 2)(x - 1)/6x

is the answer.

## Example[4x - 1/x] / (2x - 1)^{2}

To solve the given complex fraction,

first combine the numerator part 4x - 1/x.

4x = 4x^{2}/x

4x^{2}/x - 1/x

= [4x^{2} - 1]/x

4x^{2} - 1

= (2x)^{2} - 1^{2}

(2x)^{2} - 1^{2} = (2x + 1)(2x - 1)

Factor the Difference of Two Squares: a^{2} - b^{2}

4x - 1/x

= (2x + 1)(2x - 1)/x

So (given) = [(2x + 1)(2x - 1)/x] / [(2x - 1)^{2}/1].

To use the complex fraction formula,

change (2x - 1)^{2} to (2x - 1)^{2}/1.

Cancel the common factor (2x - 1)

in both of

the main numerator and the main denominator.

(2x + 1)(2x - 1)/x → (2x + 1)/x

(2x - 1)^{2}/1 → (2x - 1)/1

Solve the complex fraction.

Write the product of the outer factors,

(2x + 1)⋅1,

in the numerator.

And write the product of the inner factors,

x(2x - 1),

in the denominator.

(2x + 1)⋅1 = 2x + 1

So

(2x + 1)/[x(2x - 1)]

is the answer.