2 Multiplication and division of algebraic fractions

To multiply together two fractions (numerical or algebraic) we multiply their numerators together and then multiply their denominators together. That is

Key Point 19
Multiplication of fractions
a b × c d = a c b d

Any factors common to both numerator and denominator can be cancelled. This cancellation can be performed before or after the multiplication.

To divide one fraction by another (numerical or algebraic) we invert the second fraction and then multiply.

Key Point 20
Division of fractions
a b ÷ c d = a b × d c = a d b c b 0 , c 0 , d 0
Example 53

Simplify

  1. 2 a c × 4 c ,
  2. 2 a c × c 4 ,
  3. 2 a c ÷ 4 c
Solution
  1. 2 a c × 4 c = 8 a c 2
  2. 2 a c × c 4 = 2 a c 4 c = 2 a 4 = a 2
  3. Division is performed by inverting the second fraction and then multiplying.

    2 a c ÷ 4 c = 2 a c × c 4 = a 2  (from the result in (b))

Example 54

Simplify

  1. 1 5 x × 3 x ,
  2. 1 x × x .
Solution
  1. Note that 3 x = 3 x 1 . Then 1 5 x × 3 x = 1 5 x × 3 x 1 = 3 x 5 x = 3 5
  2. x can be written as x 1 . Then 1 x × x = 1 x × x 1 = x x = 1
Task!

Simplify

  1. 1 y × x ,
  2. y x × x .
  1. 1 y × x = 1 y × x 1 = x y
  2. y x × x = y x × x 1 = y x x = y
Example 55

Simplify 2 x y 3 x 2 y

Solution

We can write the fraction as 2 x y ÷ 3 x 2 y .

Inverting the second fraction and multiplying we find

2 x y × 2 y 3 x = 4 x y 3 x y = 4 3

Example 56

Simplify 4 x + 2 x 2 + 4 x + 3 × x + 3 7 x + 5

Solution

Factorising the numerator and denominator we find

4 x + 2 x 2 + 4 x + 3 × x + 3 7 x + 5 = 2 ( 2 x + 1 ) ( x + 1 ) ( x + 3 ) × x + 3 7 x + 5 = 2 ( 2 x + 1 ) ( x + 3 ) ( x + 1 ) ( x + 3 ) ( 7 x + 5 )

= 2 ( 2 x + 1 ) ( x + 1 ) ( 7 x + 5 )

It is usually better to factorise first and cancel any common factors before multiplying. Don’t remove any brackets unnecessarily otherwise common factors will be difficult to spot.

Task!

Simplify

15 3 x 1 ÷ 3 2 x + 1

 

To divide we invert the second fraction and multiply:

15 3 x 1 ÷ 3 2 x + 1 = 15 3 x 1 × 2 x + 1 3 = ( 5 ) ( 3 ) ( 2 x + 1 ) 3 ( 3 x 1 ) = 5 ( 2 x + 1 ) 3 x 1

Exercises
  1. Simplify
    1. 5 9 × 3 2 ,
    2. 14 3 × 3 9 ,
    3. 6 11 × 3 4 ,
    4. 4 7 × 28 3
  2. Simplify
    1. 5 9 ÷ 3 2 ,
    2. 14 3 ÷ 3 9 ,
    3. 6 11 ÷ 3 4 ,
    4. 4 7 ÷ 28 3
  3. Simplify
    1. 2 × x + y 3 ,
    2. 1 3 × 2 ( x + y ) ,
    3. 2 3 × ( x + y )
  4. Simplify
    1. 3 × x + 4 7 ,
    2. 1 7 × 3 ( x + 4 ) ,
    3. 3 7 × ( x + 4 ) ,
    4. x y × x + 1 y + 1 ,
    5. 1 y × x 2 + x y + 1 ,
    6. π d 2 4 × Q π d 2 ,
    7. Q π d 2 4
  5. Simplify 6 7 s + 3
  6. Simplify 3 x + 2 ÷ x 2 x + 4
  7. Simplify 5 2 x + 1 ÷ x 3 x 1
    1. 5 6 ,
    2. 14 9 ,
    3. 9 22 ,
    4. 16 3
    1. 10 27 ,
    2. 14 ,
    3. 8 11 ,
    4. 3 49
    1. 2 ( x + y ) 3 ,
    2. 2 ( x + y ) 3 ,
    3. 2 ( x + y ) 3
    1. 3 ( x + 4 ) 7 ,
    2. 3 ( x + 4 ) 7 ,
    3. 3 ( x + 4 ) 7 ,
    4. x ( x + 1 ) y ( y + 1 ) ,
    5. x ( x + 1 ) y ( y + 1 ) ,
    6. Q 4 ,

    7. 4 Q π d 2
  1. 6 7 ( s + 3 )
  2. 6 x
  3. 5 ( 3 x 1 ) x ( 2 x + 1 )