Skip to main content
Search IntMath

6. Multiplication and Division of Fractions

Recall the following fraction facts:

Multiplying fractions

When multiplying by a fraction, multiply numerators and multiply denominators:

`2/3xx 5/7=(2xx5)/(3xx7)=10/21`

If you can, simplify first.

In this example, we can cancel the `13` & `39` to give `1/3`:

`\cancel{13}/24xx 12/\cancel{39}=1/24 xx 12/3`

Then cancel the `12` with the `24` to give `1/2`:

`1/\cancel{24} xx \cancel{12}/3=1/2 xx 1/3`

We now multiply tops and bottoms to give:

`1/2 xx 1/3 = 1/6`

Recall: We can only multiply tops and bottoms like this. We can't add the tops and bottoms of 2 fractions, because it won't give the correct answer.

Dividing fractions

When dividing by a fraction, invert and multiply:

`3/5-:2/7=3/5xx7/2=(3xx7)/(5xx2)` `=21/10` `=2 1/10`

(I multiplied by the inverse of `2/7`, which is `7/2`.)

When we do the same things with algebraic expressions, remember to SIMPLIFY FIRST, so that the problem is easy to perform.

Example 1




Simplifying first, we cancel the 11 in the first fraction with the 33 on the bottom of the second fraction:

`\cancel{11}/5xx13/\cancel{33}` `=1/5xx13/3` `=13/15`

Example 2




We first expand out the (ax)2 on the top of the second fraction to give a2x2.

Also, because it is multiplication, we can write it as one fraction (multiply tops, multiply bottoms).

`(18sy^3)/(ax^2)xx((ax)^2)/(3s) = (18sy^3xxa^2x^2)/(ax^2xx3s)`

Next, we cancel the 18 on top with the 3 on bottom (giving 6 on top).

`(\cancel{18}sy^3xxa^2x^2)/(ax^2xx\cancel{3}s) = (6sy^3xxa^2x^2)/(ax^2xxs)`

Also we can cancel the s on top and bottom.


One a on top cancels with the a on bottom to give a on top (since a2 = a × a).

`(6y^3xxaxx\cancel{a} xx x^2)/(\cancel{a}x^2) = (6y^3xxax^2)/(x^2)`

The x2 on top cancels with the x2 on bottom to give 1.

`(6y^3xxa\cancel{x^2})/(\cancel{x^2}) = (6y^3xxa)/1`

We then multiply out what is left to give the final answer.

`= (6y^3xxa)/1=6ay^3`

Example 3




First, we invert the `(st)/4` term to give `4/(st)` and then multiply by that inverse:


Next, cancel out the two s's on the top and bottom

`(\cancel{s}r^2)/(2t)xx(4)/(\cancel{s}t) = (r^2)/(2t)xx(4)/(t)`

Then cancel the 4 on top with the 2 on bottom to give 2 on top.

`(r^2)/(\cancel{2}t)xx\cancel{4}/(t) = (r^2)/(t)xx(2)/(t)`

Then just simplfy what we are left with:





(1) `5/16-:25/13`


This one involves inverting the `25/13` to give `13/25` and then multiplying by that `13/25`.

`5/16 -: 25/13=5/16xx13/25`

Then we cancel the 5 on top with the 25 on bottom, to give `1/5`.

`=1/16 xx 13/5`


The last line is simply:

1 × 13 = 13 on top; and

16 × 5 = 80 on bottom

(2) `(9x^2-16)/(x+1)-:(4-3x)`



Dividing by `(4 − 3x)` is the same as multiplying by `1/(4 − 3x)`.

(To see why, think about this example: dividing by 2 is the same as multiplying by `1/2`.)


We factor the `(9x^2− 16)` and get `(3x + 4)(3x − 4)` using the Difference of Squares that we learned before.


We next use the following useful trick:

`(4 − 3x) = −(3x − 4)`

(To see why this works, just multiply out the right hand side.)


After cancelling, we are left with a factor of (−1) from the cancelled fraction and this negative is placed out the front for convenience.




(3) `(2x^2-18)/(x^3-25x)xx(3x-15)/(2x^2+6x)`



We need to factor everything on the top and bottom of the fractions.

For the top of the first fraction:

`2x^2− 18 = 2(x^2− 9)`

We recognise the expression in brackets to be a difference of 2 squares, which we learned about before. We can write:

`x^2− 9 = (x + 3)(x − 3)`

The bottom of the first fraction also uses difference of 2 squares:

`x^3− 25x` ` = x(x^2− 25)` ` = x(x + 5)(x −5)`

The top of the second fraction requires taking the common factor of 3 outside:

`3x − 15 = 3(x − 5)`

The bottom of the second fraction has 2x as a common factor :

`2x^2+ 6x = 2x(x + 3)`

So far we've done:

`(2x^2-18)/(x^3-25x)xx(3x-15)/(2x^2+6x) `

`=(2(x^2-9))/(x(x^2-25))xx(3(x-5))/(2x(x+3) `

`= (2(x+3)(x-3))/(x(x+5)(x-5))xx(3(x-5))/(2x(x+3)`

Next we cancel out the `2` from top and bottom:

`= (\cancel{2}(x+3)(x-3))/(x(x+5)(x-5))xx(3(x-5))/(\cancel{2}x(x+3))` `= ((x+3)(x-3))/(x(x+5)(x-5))xx(3(x-5))/(x(x+3))`

Now cancel `(x+3)` from top and bottom:

`= (\cancel{(x+3)}(x-3))/(x(x+5)(x-5))xx(3(x-5))/(x\cancel{(x+3)}` ` = ((x-3))/(x(x+5)(x-5))xx(3(x-5))/(x)`

Then cancel `(x-5)` from top and bottom:

`= ((x-3))/(x(x+5)\cancel{(x-5)})xx(3\cancel{(x-5)})/(x)` ` = ((x-3))/(x(x+5))xx(3)/(x)`

The final step is to multiply tops and multiply bottoms, since we cannot cancel anything else.




Tips, tricks, lessons, and tutoring to help reduce test anxiety and move to the top of the class.