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# Complex fractions -- Division

Back to Section 1

 A complex fraction looks like this: The numerator and/or the denominator are themselves fractions.

To simplify a complex fraction, we can immediately apply the definition of division (Lesson 6):

 ab =  a· 1b

Any fraction is equal to the numerator times the reciprocal
of the denominator.

Therefore, = pq · nm

Problem 1.   State in words how to simplify a complex fraction.

Rewrite it as the numerator times the reciprocal
of the denominator.

 Example 1.   Simplify Solution. = x2 − 25   x8 · x³ x − 5
 = (x + 5)(x − 5)        x8 · x³ x − 5 = x + 5   x5

Division -- which effectively this is -- becomes multiplication by the reciprocal. on canceling the x + 2's.

Problem 2.   Simplify.

 a) = 6 x5 · x28 = 3  4x3
 b) = 4   x − 1 · 1   x − 1 = _4_      x2 − 2x + 1
 c) = (x + 2) · x − 2x + 2 = x − 2
 d) = x + 2x + 1 · x − 2x − 1 = x2 − 4x2 − 1
 e) = −h    x(x + h) · 1h = − 1     x(x + h)

The h's cancel.  And according to the Rule of Signs, the product is negative.  (It's all right to leave the product in its factored form.)

 f) = (x + 2)(x − 2)       3x2 · _x_       (x + 2)(x + 3)
 = x − 2  3x · 1   x + 3 = x − 2   3x(x + 3)

Example 3.  If a complex fraction looks like this -- -- then we can simplify it by multiplying the numerator and denominator by c. Problem 3.   Simplify the following. = _(x + 1) − 1_(x + 1)(x − 1) = __x__x2 − 1
 Example 4.   Simplify Solution.    1-over any number is its reciprocal.  Therefore, = 43

Problem 4.   Simplify the following.

 a) = x + 1   x b) = x − 1 