# 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

Simplify

`11/5xx13/33`

**Answer:**

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

Simplify:

`(18sy^3)/(ax^2)xx((ax)^2)/(3s)`

Answer

We first expand out the (*ax*)^{2} on the top of the second fraction to give *a*^{2}*x*^{2}.

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.

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

One *a* on top cancels with the *a* on bottom to give* a* on top (since *a*^{2} = a × a).

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

The *x*^{2} on top cancels with the *x*^{2}* *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

Simplify:

`(sr^2)/(2t)-:(st)/(4)`

Answer

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

`(sr^2)/(2t)-:(st)/(4)=(sr^2)/(2t)xx(4)/(st)`

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:

`=(r^2xx2)/t^2`

`=(2r^2)/t^2`

### Exercises

Simplify:

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

Answer

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`

`=13/80`

The last line is simply:

1 × 13 = 13 on top; and

16 × 5 = 80 on bottom

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

Answer

`(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`.)

`(9x^2-16)/(x+1)xx1/(4-3x)`

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

`((3x+4)(3x-4))/(x+1)xx1/(4-3x)`

We next use the following useful trick:

`(4 − 3x) = −(3x − 4)`

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

`((3x+4)(3x-4))/(x+1)xx1/-(3x-4)`

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

`=-(3x+4)/(x+1)`

So

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

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

Answer

`(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 2*x* 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.

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

So

`(2x^2-18)/(x^3-25x)xx(3x-15)/(2x^2+6x)=(3(x-3))/(x^2(x+5))`

### Search IntMath

### Online Algebra Solver

This algebra solver can solve a wide range of math problems.

Go to: Online algebra solver