Functions and algebra: Manipulate and simplify algebraic expressions

# Unit 1: Multiplying algebraic fractions

Dylan Busa

### Unit outcomes

By the end of this unit you will be able to:

• Multiply algebraic fractions using factorisation.
• Divide algebraic fractions using factorisation.

## What you should know

Before you start this unit, make sure you can:

• Manipulate and simplify algebraic expressions. Refer to level 2 subject outcome 2.2 units 1 and 3 if you need help with this.
• Simplify algebraic expressions by means of factorisation. Refer to level 2 subject outcome 2.2 unit 2 if you need help with this.

Here is a short self-assessment to make sure you have the skills you need to proceed with this unit:

1. Simplify $\scriptsize 2(3a+b)(3a-b)-{{(3a-b)}^{2}}$.
2. Factorise $\scriptsize {{(x+y)}^{2}}-16{{y}^{2}}$.
3. Simplify $\scriptsize \displaystyle \frac{{9{{a}^{2}}-4{{b}^{2}}}}{{18ab}}\times \displaystyle \frac{{6a}}{{3a-2b}}$.

Solutions

1. .
\scriptsize \begin{align*}2(3a+b)(3a-b)-{{(3a-b)}^{2}}&=2(9{{a}^{2}}-{{b}^{2}})-(9{{a}^{2}}-6ab+{{b}^{2}})\\&=18{{a}^{2}}-2{{b}^{2}}-9{{a}^{2}}+6ab-{{b}^{2}}\\&=9{{a}^{2}}+6ab-3{{b}^{2}}\end{align*}
2. .
\scriptsize \begin{align*}{{(x+y)}^{2}}-16{{y}^{2}}&=\left( {\sqrt{{{{{(x+y)}}^{2}}}}-\sqrt{{16{{y}^{2}}}}} \right)\left( {\sqrt{{{{{(x+y)}}^{2}}}}+\sqrt{{16{{y}^{2}}}}} \right)\\&=(x+y-4y)(x+y+4y)\\&=(x-3y)(x+5y)\end{align*}
3. .
\scriptsize \begin{align*}\displaystyle \frac{{9{{a}^{2}}-4{{b}^{2}}}}{{18ab}}\times \displaystyle \frac{{6a}}{{3a-2b}}&=\displaystyle \frac{{(3a-2b)(3a+2b)}}{{18ab}}\times \displaystyle \frac{{6a}}{{(3a-2b)}}\\&=\displaystyle \frac{{(3a+2b)}}{{3\text{ }\cancel{1}\cancel{8}\cancel{a}b}}\times \displaystyle \frac{{\cancel{6}\cancel{a}}}{1}&& \text{Factors of }(3a-2b)\text{ cancel}\\&=\displaystyle \frac{{3a+2b}}{{3b}}\end{align*}

## Introduction

Algebraic expressions are the bread and butter of much of mathematics. To mix metaphors, they are the nuts and bolts of equations. In order to solve equations, you need to be able to simplify them and the algebraic expressions from which they are made.

Remember that an algebraic expression is made up of constants and variables joined together by addition, subtraction, multiplication and division.
In the expression $\scriptsize 3x+2$:

• the $\scriptsize 2$ is a constant because its value does not change
• the $\scriptsize x$ is a variable because its value can change
• the $\scriptsize 3$ in front of the variable $\scriptsize x$ is called the coefficient of $\scriptsize x$ and is being multiplied by the $\scriptsize x$
• the $\scriptsize 3x$ and $\scriptsize 2$ are called the terms of the algebraic expression. Terms are separated by addition or subtraction. In this algebraic expression we have two terms.

We usually have to simplify algebraic expressions to make them easier to work with or calculate with or use in some other way.

## Factorisation recap

Before we dive in, let’s spend some time revising what we know about factorisation.

In the same way that we can split any number into its factors (numbers that multiply together to produce the original number) we can also factorise algebraic expressions. For example, the factors of $\scriptsize 12$, in their pairs, are $\scriptsize 1\text{ and }12$; $\scriptsize 2\text{ and }6$; $\scriptsize 3\text{ and }4$. These are just the whole number factor pairs. $\scriptsize \displaystyle \frac{1}{2}\text{ and }24$ are also factors of $\scriptsize 12$ because $\scriptsize \displaystyle \frac{1}{2}\times 24=12$.

Similarly, we can say that the factors of $\scriptsize 4{{x}^{2}}$ are $\scriptsize 4\text{ and }x\text{ and }x$ because $\scriptsize 4\times x\times x=4{{x}^{2}}$.

### Factorising by taking out the highest common factor

Taking out the (usually) highest common factor is the first step of factorising algebraic expressions. It is the simplest to do but is also the first thing you should try when you need to split an algebraic fraction into its factors (i.e. factorise it). Here is an example:

$\scriptsize 4{{x}^{3}}+2{{x}^{2}}+6x$

We can see the common factors in each term more easily if we expand each term.

$\scriptsize \left( {2\times 2\times x\times x\times x} \right)+\left( {2\times x\times x} \right)+\left( {2\times 3\times x} \right)$

The highest common factor we can take out of each term is $\scriptsize 2\times x$.

$\scriptsize \left( {2\times 2\times x\times x\times x} \right)+\left( {2\times x\times x} \right)+\left( {2\times 3\times x} \right)$

If we take this highest common factor out of each term we get:

\scriptsize \begin{align*}&4{{x}^{3}}+2{{x}^{2}}+6x\\&=2x\left[ {\left( {2\times x\times x} \right)+\left( x \right)+\left( 3 \right)} \right]\\&=2x\left( {2{{x}^{2}}+x+3} \right)\end{align*}

### Factorising by grouping

Factorising by grouping is another way to create highest common factors in different parts of an algebraic expression. This is done by grouping these different parts together. Here is an example:

$\scriptsize 3ax+bx-3ay-by-9a-3b=\left( {3ax+bx} \right)-\left( {3ay+by} \right)-\left( {9a+3b} \right)$

You need to be careful of the signs inside the brackets. If you multiply the brackets out, you must always be left with the original expression.

Now we can take a common factor out of each bracket:

\scriptsize \begin{align*}3ax+bx-3ay-by-9a-3b&=\left( {3ax+bx} \right)-\left( {3ay+by} \right)-\left( {9a+3b} \right)\\ &=x(3a+b)-y(3a+b)-3(3a+b)\end{align*}

Can you see that we now have a common factor in each term of $\scriptsize (3a+b)$? Let’s take it out:

\scriptsize \begin{align*}3ax+bx-3ay-by-9a-3b&=\left( {3ax+bx} \right)-\left( {3ay+by} \right)-\left( {9a+3b} \right)\\ &=x(3a+b)-y(3a+b)-3(3a+b)\\ &=(3a+b)(x-y-3)\end{align*}

### Factorising the difference of two squares

This factorisation technique only applies to algebraic expressions where there are two terms subtracted from each other. The pattern is always the same.

$\scriptsize {{x}^{2}}-{{y}^{2}}=(x+y)(x-y)$

Sometimes, as in the next example, you need to do some extra work before you can get to the difference of two squares.

\scriptsize \begin{align*}&2a({{a}^{2}}-9)-7({{a}^{2}}-9)&&\text{Take out the common factor of }({{a}^{2}}-9)\\ &=({{a}^{2}}-9)(2a-7)&&\text{Now factorise the difference of two squares}\\&=(a+3)(a-3)(2a-7)&& \text{Don't forget the original common factor of }(2a-7)\end{align*}

You don’t need to have perfect squares to factorise using this method. Just follow the basic pattern:

$\scriptsize x-y=\left( {\sqrt{x}+\sqrt{y}} \right)\left( {\sqrt{x}-\sqrt{y}} \right)$

A trinomial is an algebraic expression with three terms. A quadratic trinomial is an algebraic expression of the form $\scriptsize a{{x}^{2}}+bx+c$.

In level 2, we saw how to factorise quadratic trinomials using trial and error or a method called grouping. You can use whichever method you find easiest. Her are two examples:

\scriptsize \begin{align*}&2{{y}^{2}}+5y-3&&\text{This quadratic trinomial is in the standard form aleady}\\ &=(2y+1)(y-3)&& \text{Multiply the binomial together to make sure you get the original}\end{align*}

\scriptsize \begin{align*}&7x-6+5{{x}^{2}}&&\text{This quadratic trinomial is not yet in standard form}\\ &=5{{x}^{2}}+7x-6\\ &=(5x-3)(x+2)&& \text{Multiply the binomial together to make sure you get the original expression}\end{align*}

## Multiply algebraic fractions using factorisation

Multiplying algebraic fractions together works the same way as multiplying any other fractions together. We multiply the numerators to find the numerator of the product, and then we multiply the denominators to find the denominator of the product.

However, if we factorise the numerators and denominators of the fractions, we may find factors that we can ‘cancel’, making the multiplication of numerators and denominators simpler.

Remember, we never actually cancel anything. If we have the same factor in the numerator and the denominator, we recognise that the value of this is equal to $\scriptsize 1$ and therefore it has no influence on the value of the final product.

### Example 1.1

Multiply the algebraic fractions and simplify:

$\scriptsize \displaystyle \frac{{{{x}^{2}}+4x-5}}{{3x+18}}\times \displaystyle \frac{{x+6}}{{x+5}}$

Solution

The first thing we need to do is factorise as many of the numerators and denominators as we can:

\scriptsize \begin{align*}&\displaystyle \frac{{{{x}^{2}}+4x-5}}{{3x+18}}\times \displaystyle \frac{{x+6}}{{x+5}}\\&=\displaystyle \frac{{(x-1)(x+5)}}{{3(x+6)}}\times \displaystyle \frac{{x+6}}{{x+5}}\end{align*}

Now you can either multiply the numerators and denominators and then ‘cancel’, or ‘cancel’ and then multiply the numerators and denominators.

\scriptsize \begin{align*}&\displaystyle \frac{{(x-1)(x+5)}}{{3(x+6)}}\times \displaystyle \frac{{x+6}}{{x+5}}\\&=\displaystyle \frac{{(x-1)\bcancel{{(x+5)}}\bcancel{{(x+6)}}}}{{3\bcancel{{(x+6)}}\bcancel{{(x+5)}}}}\\&=\displaystyle \frac{{(x-1)}}{3}\end{align*}
OR
\scriptsize \begin{align*}&\displaystyle \frac{{(x-1)\bcancel{{(x+5)}}}}{{3\bcancel{{(x+6)}}}}\times \displaystyle \frac{{\bcancel{{x+6}}}}{{\bcancel{{x+5}}}}\\&=\displaystyle \frac{{(x-1)}}{3}\end{align*}

### Example 1.2

Simplify the algebraic expression:

$\scriptsize \displaystyle \frac{{{{x}^{2}}+11x+30}}{{{{x}^{2}}+5x+6}}\cdot \displaystyle \frac{{{{x}^{2}}+7x+12}}{{{{x}^{2}}+8x+16}}$

Solution

\scriptsize \begin{align*}&\displaystyle \frac{{{{x}^{2}}+11x+30}}{{{{x}^{2}}+5x+6}}\cdot \displaystyle \frac{{{{x}^{2}}+7x+12}}{{{{x}^{2}}+8x+16}}\\&=\displaystyle \frac{{(x+5)(x+6)}}{{(x+2)\bcancel{{(x+3)}}}}\cdot \displaystyle \frac{{\bcancel{{(x+4)}}\bcancel{{(x+3)}}}}{{\bcancel{{(x+4)}}(x+4)}}\\&=\displaystyle \frac{{(x+5)(x+6)}}{{(x+2)(x+4)}}\end{align*}

Note: While we could multiply the numerator and denominator out, we tend to leave them in factorised form. This is usually considered the simplest form of an expression.

### Take note!

If you have an expression such as $\scriptsize \displaystyle \frac{{x+3}}{{x+3}}$, you can ‘cancel’ the whole factor of $\scriptsize (x+3)$ in the numerator with the whole factor of $\scriptsize (x+3)$ in the denominator.
$\scriptsize \displaystyle \frac{{x+3}}{{x+3}}=\displaystyle \frac{{\bcancel{{(x+3)}}}}{{\bcancel{{(x+3)}}}}=1$

If you have an expression such as $\scriptsize \displaystyle \frac{{x+3}}{{x-3}}$, you cannot ‘cancel’ the $\scriptsize x$’s or the $\scriptsize 3$‘s because these are separate terms separated by $\scriptsize +$ and $\scriptsize -$ signs.

$\scriptsize \displaystyle \frac{{x+3}}{{x-3}}\ne \displaystyle \frac{{\bcancel{x}+\bcancel{3}}}{{\bcancel{x}-\bcancel{3}}}$

Convince yourself that this is true by substituting different values for $\scriptsize x$.

### Exercise 1.1

Simplify the following algebraic expressions:

1. $\scriptsize \displaystyle \frac{{n-12}}{{{{n}^{2}}-144}}$
2. $\scriptsize \displaystyle \frac{{{{x}^{2}}-x-6}}{{{{x}^{2}}-9}}\cdot \displaystyle \frac{{2{{x}^{2}}+7x-15}}{{2{{x}^{2}}+x-6}}$
3. $\scriptsize \displaystyle \frac{2}{{{{{(x-y)}}^{2}}}}\times \displaystyle \frac{{{{x}^{2}}-{{y}^{2}}}}{6}$
4. $\scriptsize \displaystyle \frac{{{{t}^{2}}-1}}{{{{t}^{2}}+4t+3}}\cdot \displaystyle \frac{{{{t}^{2}}+2t-15}}{{{{t}^{2}}-4t+3}}$
5. $\scriptsize \displaystyle \frac{{36{{x}^{2}}-25}}{{5{{x}^{2}}+70x+120}}\cdot \displaystyle \frac{{3{{x}^{2}}+36x}}{{2{{x}^{2}}-18}}\cdot \displaystyle \frac{{{{x}^{2}}-2x-3}}{{6{{x}^{2}}+11x+5}}$

The full solutions are at the end of the unit.

## Divide algebraic fractions using factorisation

Dividing algebraic fractions works the same way as the division of other fractions. To divide one algebraic fraction by another one, multiply the first expression by the reciprocal of the second. In other words:

$\scriptsize \displaystyle \frac{1}{x}\div \displaystyle \frac{{{{x}^{2}}}}{3}=\displaystyle \frac{1}{x}\times \displaystyle \frac{3}{{{{x}^{2}}}}$

Once the division expression has been rewritten as a multiplication expression, we can multiply the fractions as normal.

### Example 1.3

Divide the following algebraic expression and simplify:

$\scriptsize \displaystyle \frac{{2{{x}^{2}}+x-6}}{{{{x}^{2}}-1}}\div \displaystyle \frac{{{{x}^{2}}-4}}{{{{x}^{2}}+2x+1}}$

Solution

The first step is to rewrite the division expression as a multiplication expression.

\scriptsize \begin{align*}&\displaystyle \frac{{2{{x}^{2}}+x-6}}{{{{x}^{2}}-1}}\div \displaystyle \frac{{{{x}^{2}}-4}}{{{{x}^{2}}+2x+1}}\\&=\displaystyle \frac{{2{{x}^{2}}+x-6}}{{{{x}^{2}}-1}}\times \displaystyle \frac{{{{x}^{2}}+2x+1}}{{{{x}^{2}}-4}}\end{align*}

Now we can proceed as normal.

\scriptsize \begin{align*}&\displaystyle \frac{{2{{x}^{2}}+x-6}}{{{{x}^{2}}-1}}\times \displaystyle \frac{{{{x}^{2}}+2x+1}}{{{{x}^{2}}-4}}\\&=\displaystyle \frac{{(2x-3)\bcancel{{(x+2)}}}}{{\bcancel{{(x+1)}}(x-1)}}\times \displaystyle \frac{{\bcancel{{(x+1)}}(x+1)}}{{\bcancel{{(x+2)}}(x-2)}}\\&=\displaystyle \frac{{(2x-3)(x+1)}}{{(x-1)(x-2)}}\end{align*}

### Example 1.4

Divide the following algebraic expression and simplify:

$\scriptsize \displaystyle \frac{{9{{x}^{2}}-16}}{{3{{x}^{2}}+25x+28}}\div \displaystyle \frac{{3{{x}^{2}}-10x+8}}{{{{x}^{2}}+5x-14}}$

Solution

Rewrite the division expression as a multiplication expression by multiplying by the reciprocal of the second fraction.

\scriptsize \begin{align*}&\displaystyle \frac{{9{{x}^{2}}-16}}{{3{{x}^{2}}+25x+28}}\div \displaystyle \frac{{3{{x}^{2}}-10x+8}}{{{{x}^{2}}+5x-14}}\\&=\displaystyle \frac{{9{{x}^{2}}-16}}{{3{{x}^{2}}+25x+28}}\times \displaystyle \frac{{{{x}^{2}}+5x-14}}{{3{{x}^{2}}-10x+8}}\end{align*}

Now proceed as normal to factorise the numerators and denominators.

\scriptsize \begin{align*}&\displaystyle \frac{{9{{x}^{2}}-16}}{{3{{x}^{2}}+25x+28}}\times \displaystyle \frac{{{{x}^{2}}+5x-14}}{{3{{x}^{2}}-10x+8}}\\&\text{=}\displaystyle \frac{{\bcancel{{(3x+4)}}\bcancel{{(3x-4)}}}}{{\bcancel{{(3x+4)}}\bcancel{{(x+7)}}}}\times \displaystyle \frac{{\bcancel{{(x-2)}}\bcancel{{(x+7)}}}}{{\bcancel{{(3x-4)}}\bcancel{{(x-2)}}}}\\&=1\end{align*}

### Note

The reciprocal of a number is its multiplicative inverse. If $\scriptsize x$ is any non-zero number, then its reciprocal is $\scriptsize \displaystyle \frac{1}{x}$ such that $\scriptsize x\times \displaystyle \frac{1}{x}=1$.

### Exercise 1.2

Divide the following algebraic expressions and simplify.

1. $\scriptsize \displaystyle \frac{{2{{x}^{2}}+x-3}}{{7{{x}^{2}}-7}}\div \displaystyle \frac{{2{{x}^{2}}-x-6}}{{3x+3}}$
2. $\scriptsize \displaystyle \frac{{{{q}^{2}}-9}}{{{{q}^{2}}+6q+9}}\div \displaystyle \frac{{{{q}^{2}}-2q-3}}{{{{q}^{2}}+2q-3}}$
3. $\scriptsize \displaystyle \frac{{144{{b}^{2}}-25}}{{72{{b}^{2}}-6b-10}}\div \displaystyle \frac{{24{{b}^{2}}+22b+5}}{{36{{b}^{2}}-9}}$

The full solutions are at the end of the unit.

### Simplify complex fractions

Let’s now look at how to simplify complex fractions.

### Example 1.5

Simplify the algebraic expression:

$\scriptsize \displaystyle \frac{{{{x}^{2}}+x-6}}{{{{x}^{2}}-2x-3}}\cdot \displaystyle \frac{{2{{x}^{2}}-3x-9}}{{{{x}^{2}}-x-2}}\div \displaystyle \frac{{10{{x}^{2}}+27x+18}}{{{{x}^{2}}+2x+1}}$

Solution

In this case, we can either multiply the first two fractions and then deal with the division of the third fraction or we can rewrite the division as a multiplication straight away. Rewriting it as a multiplication straight away will allow us to ‘cancel’ in one step.

\scriptsize \begin{align*}&\displaystyle \frac{{{{x}^{2}}+x-6}}{{{{x}^{2}}-2x-3}}\cdot \displaystyle \frac{{2{{x}^{2}}-3x-9}}{{{{x}^{2}}-x-2}}\div \displaystyle \frac{{10{{x}^{2}}+27x+18}}{{{{x}^{2}}+2x+1}}\\&=\displaystyle \frac{{{{x}^{2}}+x-6}}{{{{x}^{2}}-2x-3}}\cdot \displaystyle \frac{{2{{x}^{2}}-3x-9}}{{{{x}^{2}}-x-2}}\times \displaystyle \frac{{{{x}^{2}}+2x+1}}{{10{{x}^{2}}+27x+18}}\\&=\displaystyle \frac{{\bcancel{{(x-2)}}(x+3)}}{{\bcancel{{(x+1)}}\bcancel{{(x-3)}}}}\times \displaystyle \frac{{\bcancel{{(2x+3)}}\bcancel{{(x-3)}}}}{{\bcancel{{(x+1)}}\bcancel{{(x-2)}}}}\times \displaystyle \frac{{\bcancel{{(x+1)}}\bcancel{{(x+1)}}}}{{(5x+6)\bcancel{{(2x+3)}}}}\\&=\displaystyle \frac{{x+3}}{{5x+6}}\end{align*}

### Example 1.6

Simplify the algebraic expression:

$\scriptsize \displaystyle \frac{{{{x}^{2}}+7x+12}}{{{{x}^{2}}+x-6}}\div \displaystyle \frac{{{{x}^{2}}-16}}{{{{x}^{2}}+4x+3}}\div \displaystyle \frac{{3x+9}}{{3{{x}^{2}}-12}}$

Solution

When you have an expression with two division signs, you can rewrite both of them by multiplying by the reciprocal of each fraction.

\scriptsize \begin{align*}&\displaystyle \frac{{{{x}^{2}}+7x+12}}{{{{x}^{2}}+x-6}}\div \displaystyle \frac{{{{x}^{2}}-16}}{{{{x}^{2}}+4x+3}}\div \displaystyle \frac{{3x+9}}{{3{{x}^{2}}-12}}\\&=\displaystyle \frac{{{{x}^{2}}+7x+12}}{{{{x}^{2}}+x-6}}\times \displaystyle \frac{{{{x}^{2}}+4x+3}}{{{{x}^{2}}-16}}\times \displaystyle \frac{{3{{x}^{2}}-12}}{{3x+9}}\\&=\displaystyle \frac{{(x+3)(x+4)}}{{(x-2)(x+3)}}\times \displaystyle \frac{{(x+1)(x+3)}}{{(x+4)(x-4)}}\times \displaystyle \frac{{3({{x}^{2}}-4)}}{{3(x+3)}}\\&=\displaystyle \frac{{\bcancel{{(x+3)}}\bcancel{{(x+4)}}}}{{\bcancel{{(x-2)}}\bcancel{{(x+3)}}}}\times \displaystyle \frac{{(x+1)\bcancel{{(x+3)}}}}{{\bcancel{{(x+4)}}(x-4)}}\times \displaystyle \frac{{\bcancel{3}(x+2)\bcancel{{(x-2)}}}}{{\bcancel{3}\bcancel{{(x+3)}}}}\\&=\displaystyle \frac{{(x+1)(x+2)}}{{(x-4)}}\end{align*}

### Exercise 1.3

Simplify the following:

1. $\scriptsize \displaystyle \frac{{{{x}^{2}}+6x-7}}{{x+1}}\cdot \displaystyle \frac{{{{x}^{2}}-x-2}}{{2x-14}}\div \displaystyle \frac{{5{{x}^{3}}-20x}}{6}$
2. $\scriptsize \left( {\displaystyle \frac{{2{{x}^{2}}-3x-9}}{{{{x}^{2}}-9}}} \right)\left( {\displaystyle \frac{{2{{x}^{3}}+8{{x}^{2}}+6x}}{{16{{x}^{3}}+24{{x}^{2}}}}} \right)\div \displaystyle \frac{{{{x}^{2}}-x-2}}{{{{x}^{2}}+3x-10}}$
3. $\scriptsize \displaystyle \frac{{\left( {\displaystyle \frac{{3{{y}^{2}}-10y+3}}{{3{{y}^{2}}+5y-2}}\cdot \displaystyle \frac{{2{{y}^{2}}-3y-20}}{{2{{y}^{2}}-y-15}}} \right)}}{{y-4}}$

The full solutions are at the end of the unit.

## Summary

In this unit you have learnt the following:

• How to multiply algebraic fractions by first factorising the numerators and denominators in order to ‘cancel’ like factors.
• How to divide algebraic fractions by multiplying by the reciprocal of the fraction.

# Unit 1: Assessment

#### Suggested time to complete: 25 minutes

Simplify the following:

1. $\scriptsize \displaystyle \frac{{{{a}^{2}}+9a+18}}{{{{a}^{2}}+3a-18}}$
2. $\scriptsize \displaystyle \frac{{{{c}^{2}}+2c-24}}{{{{c}^{2}}+12c+36}}\cdot \displaystyle \frac{{{{c}^{2}}-10c+24}}{{{{c}^{2}}-8c+16}}$
3. $\scriptsize \displaystyle \frac{{16{{a}^{2}}-24a+9}}{{4{{a}^{2}}+17a-15}}\div \displaystyle \frac{{16{{a}^{2}}-9}}{{4{{a}^{2}}+11a+6}}$
4. $\scriptsize \displaystyle \frac{{2{{x}^{2}}+x}}{{2x-1}}\times \displaystyle \frac{{4{{x}^{2}}-1}}{{{{x}^{2}}}}\div \displaystyle \frac{{2x+1}}{x}$
5. $\scriptsize \displaystyle \frac{{3{{x}^{2}}-21x}}{{18{{x}^{2}}+24x}}\div \displaystyle \frac{{x-7}}{{4x+5}}\cdot \displaystyle \frac{{18{{x}^{3}}-32x}}{{4{{x}^{2}}+x-5}}\div \displaystyle \frac{{3x-4}}{{x-1}}$

The full solutions are at the end of the unit.

# Unit 1: Solutions

### Exercise 1.1

1. .
\scriptsize \begin{align*}&\displaystyle \frac{{n-12}}{{{{n}^{2}}-144}}\\&=\displaystyle \frac{{\bcancel{{n-12}}}}{{(n+12)\bcancel{{(n-12)}}}}\\&=\displaystyle \frac{1}{{n+12}}\end{align*}
2. .
\scriptsize \begin{align*}&\displaystyle \frac{{{{x}^{2}}-x-6}}{{{{x}^{2}}-9}}\cdot \displaystyle \frac{{2{{x}^{2}}+7x-15}}{{2{{x}^{2}}+x-6}}\\&=\displaystyle \frac{{\bcancel{{(x+2)}}\bcancel{{(x-3)}}}}{{(x+3)\bcancel{{(x-3)}}}}\cdot \displaystyle \frac{{\bcancel{{(2x-3)}}(x+5)}}{{\bcancel{{(2x-3)}}\bcancel{{(x+2)}}}}\\&=\displaystyle \frac{{(x+5)}}{{(x+3)}}\end{align*}
3. .
\scriptsize \begin{align*}&\displaystyle \frac{2}{{{{{(x-y)}}^{2}}}}\times \displaystyle \frac{{{{x}^{2}}-{{y}^{2}}}}{6}\\&=\displaystyle \frac{{\bcancel{2}}}{{\bcancel{{(x-y)}}(x-y)}}\times \displaystyle \frac{{(x+y)\bcancel{{(x-y)}}}}{{\bcancel{6}3}}\\&=\displaystyle \frac{{(x+y)}}{{3(x-y)}}\end{align*}
4. .
\scriptsize \begin{align*}&\displaystyle \frac{{{{t}^{2}}-1}}{{{{t}^{2}}+4t+3}}\cdot \displaystyle \frac{{{{t}^{2}}+2t-15}}{{{{t}^{2}}-4t+3}}\\&=\displaystyle \frac{{\bcancel{{(t+1)}}\bcancel{{(t-1)}}}}{{\bcancel{{(t+1)}}(t+3)}}\cdot \displaystyle \frac{{\bcancel{{(t-3)}}(t+5)}}{{\bcancel{{(t-1)}}\bcancel{{(t-3)}}}}\\&=\displaystyle \frac{{(t+5)}}{{(t+3)}}\end{align*}
5. .
\scriptsize \begin{align*}&\displaystyle \frac{{36{{x}^{2}}-25}}{{5{{x}^{2}}+70x+120}}\cdot \displaystyle \frac{{3{{x}^{2}}+36x}}{{2{{x}^{2}}-18}}\cdot \displaystyle \frac{{{{x}^{2}}-2x-3}}{{6{{x}^{2}}+11x+5}}\\&=\displaystyle \frac{{\bcancel{{(6x+5)}}(6x-5)}}{{5\bcancel{{(x+12)}}(x+2)}}\cdot \displaystyle \frac{{3x\bcancel{{(x+12)}}}}{{2(x+3)\bcancel{{(x-3)}}}}\cdot \displaystyle \frac{{\bcancel{{(x+1)}}\bcancel{{(x-3)}}}}{{\bcancel{{(6x+5)}}\bcancel{{(x+1)}}}}\\&=\displaystyle \frac{{3x(6x-5)}}{{10(x+2)(x+3)}}\end{align*}

Back to Exercise 1.1

### Exercise 1.2

1. .
\scriptsize \begin{align*}&\displaystyle \frac{{2{{x}^{2}}+x-3}}{{7{{x}^{2}}-7}}\div \displaystyle \frac{{2{{x}^{2}}-x-6}}{{3x+3}}\\&=\displaystyle \frac{{2{{x}^{2}}+x-3}}{{7{{x}^{2}}-7}}\times \displaystyle \frac{{3x+3}}{{2{{x}^{2}}-x-6}}\\&=\displaystyle \frac{{\bcancel{{(2x+3)}}\bcancel{{(x-1)}}}}{{7\bcancel{{(x+1)}}\bcancel{{(x-1)}}}}\times \displaystyle \frac{{3\bcancel{{(x+1)}}}}{{\bcancel{{(2x+3)}}(x-2)}}\\&=\displaystyle \frac{3}{{7(x-2)}}\end{align*}
2. .
\scriptsize \begin{align*}&\displaystyle \frac{{{{q}^{2}}-9}}{{{{q}^{2}}+6q+9}}\div \displaystyle \frac{{{{q}^{2}}-2q-3}}{{{{q}^{2}}+2q-3}}\\&=\displaystyle \frac{{{{q}^{2}}-9}}{{{{q}^{2}}+6q+9}}\times \displaystyle \frac{{{{q}^{2}}+2q-3}}{{{{q}^{2}}-2q-3}}\\&=\displaystyle \frac{{\bcancel{{(q+3)}}\bcancel{{(q-3)}}}}{{\bcancel{{(q+3)}}\bcancel{{(q+3)}}}}\times \displaystyle \frac{{(q-1)\bcancel{{(q+3)}}}}{{(q+1)\bcancel{{(q-3)}}}}\\&=\displaystyle \frac{{(q-1)}}{{(q+1)}}\end{align*}
3. .
\scriptsize \begin{align*}&\displaystyle \frac{{144{{b}^{2}}-25}}{{72{{b}^{2}}-6b-10}}\div \displaystyle \frac{{24{{b}^{2}}+22b+5}}{{36{{b}^{2}}-9}}\\&=\displaystyle \frac{{144{{b}^{2}}-25}}{{72{{b}^{2}}-6b-10}}\times \displaystyle \frac{{36{{b}^{2}}-9}}{{24{{b}^{2}}+22b+5}}\\&=\displaystyle \frac{{\bcancel{{(12b+5)}}\bcancel{{(12b-5)}}}}{{2\bcancel{{(12b-5)}}(3b+1)}}\times \displaystyle \frac{{9\bcancel{{(2b+1)}}(2b-1)}}{{\bcancel{{(12b+5)}}\bcancel{{(2b+1)}}}}\\&=\displaystyle \frac{{9(2b-1)}}{{2(3b+1)}}\end{align*}

Back to Exercise 1.2

### Exercise 1.3

1. .
\scriptsize \begin{align*}&\displaystyle \frac{{{{x}^{2}}+6x-7}}{{x+1}}\cdot \displaystyle \frac{{{{x}^{2}}-x-2}}{{2x-14}}\div \displaystyle \frac{{5{{x}^{3}}-20x}}{6}\\&=\displaystyle \frac{{{{x}^{2}}+6x-7}}{{x+1}}\cdot \displaystyle \frac{{{{x}^{2}}-x-2}}{{2x-14}}\times \displaystyle \frac{6}{{5{{x}^{3}}-20x}}\\&=\displaystyle \frac{{{{x}^{2}}+6x-7}}{{x+1}}\cdot \displaystyle \frac{{{{x}^{2}}-x-2}}{{2x-14}}\times \displaystyle \frac{6}{{5x({{x}^{2}}-4)}}\\&=\displaystyle \frac{{\bcancel{{(x+1)}}\bcancel{{(x-7)}}}}{{\bcancel{{(x+1)}}}}\cdot \displaystyle \frac{{(x+1)\bcancel{{(x-2)}}}}{{2\bcancel{{(x-7)}}}}\times \displaystyle \frac{6}{{5x(x+2)\bcancel{{(x-2)}}}}\\&=\displaystyle \frac{{6(x+1)}}{{10x(x+2)}}\\&=\displaystyle \frac{{3(x+1)}}{{5x(x+2)}}\end{align*}
2. .
\scriptsize \begin{align*}&\displaystyle \frac{{2{{x}^{2}}-3x-9}}{{{{x}^{2}}-9}}\times \displaystyle \frac{{2{{x}^{3}}+8{{x}^{2}}+6x}}{{16{{x}^{3}}+24{{x}^{2}}}}\div \displaystyle \frac{{{{x}^{2}}-x-2}}{{{{x}^{2}}+3x-10}}\\&=\displaystyle \frac{{2{{x}^{2}}-3x-9}}{{{{x}^{2}}-9}}\times \displaystyle \frac{{2{{x}^{3}}+8{{x}^{2}}+6x}}{{16{{x}^{3}}+24{{x}^{2}}}}\times \displaystyle \frac{{{{x}^{2}}+3x-10}}{{{{x}^{2}}-x-2}}\\&=\displaystyle \frac{{2{{x}^{2}}-3x-9}}{{{{x}^{2}}-9}}\times \displaystyle \frac{{2x({{x}^{2}}+4x+3)}}{{16{{x}^{3}}+24{{x}^{2}}}}\times \displaystyle \frac{{{{x}^{2}}+3x-10}}{{{{x}^{2}}-x-2}}\\&=\displaystyle \frac{{\bcancel{{(2x+3)}}\bcancel{{(x-3)}}}}{{\bcancel{{(x+3)}}\bcancel{{(x-3)}}}}\times \displaystyle \frac{{2x\bcancel{{(x+3)}}\bcancel{{(x+1)}}}}{{8{{x}^{2}}\bcancel{{(2x+3)}}}}\times \displaystyle \frac{{\bcancel{{(x-2)}}(x+5)}}{{\bcancel{{(x+1)}}\bcancel{{(x-2)}}}}\\&=\displaystyle \frac{{2x(x+5)}}{{8{{x}^{2}}}}\\&=\displaystyle \frac{{(x+5)}}{{4x}}\end{align*}
3. .
\scriptsize \begin{align*}&\displaystyle \frac{{\left( {\displaystyle \frac{{3{{y}^{2}}-10y+3}}{{3{{y}^{2}}+5y-2}}\cdot \displaystyle \frac{{2{{y}^{2}}-3y-20}}{{2{{y}^{2}}-y-15}}} \right)}}{{y-4}}\\&=\left( {\displaystyle \frac{{3{{y}^{2}}-10y+3}}{{3{{y}^{2}}+5y-2}}\cdot \displaystyle \frac{{2{{y}^{2}}-3y-20}}{{2{{y}^{2}}-y-15}}} \right)\div \displaystyle \frac{{y-4}}{1}\\&=\left( {\displaystyle \frac{{3{{y}^{2}}-10y+3}}{{3{{y}^{2}}+5y-2}}\cdot \displaystyle \frac{{2{{y}^{2}}-3y-20}}{{2{{y}^{2}}-y-15}}} \right)\times \displaystyle \frac{1}{{y-4}}\\&=\displaystyle \frac{{\bcancel{{(3y-1)}}\bcancel{{(y-3)}}}}{{\bcancel{{(3y-1)}}(y+2)}}\cdot \displaystyle \frac{{\bcancel{{(2y+5)}}\bcancel{{(y-4)}}}}{{\bcancel{{(2y+5)}}\bcancel{{(y-3)}}}}\times \displaystyle \frac{1}{{\bcancel{{(y-4)}}}}\\&=\displaystyle \frac{1}{{(y+2)}}\end{align*}

Back to Exercise 1.3

### Unit 1: Assessment

1. .
\scriptsize \begin{align*}&\displaystyle \frac{{{{a}^{2}}+9a+18}}{{{{a}^{2}}+3a-18}}\\&=\displaystyle \frac{{(a+3)\bcancel{{(a+6)}}}}{{(a-3)\bcancel{{(a+6)}}}}\\&=\displaystyle \frac{{a+3}}{{a-3}}\end{align*}
2. .
\scriptsize \begin{align*}&\displaystyle \frac{{{{c}^{2}}+2c-24}}{{{{c}^{2}}+12c+36}}\cdot \displaystyle \frac{{{{c}^{2}}-10c+24}}{{{{c}^{2}}-8c+16}}\\&=\displaystyle \frac{{\bcancel{{(c-4)}}\bcancel{{(c+6)}}}}{{(c+6)\bcancel{{(c+6)}}}}\cdot \displaystyle \frac{{(c-6)\bcancel{{(c-4)}}}}{{\bcancel{{(c-4)}}\bcancel{{(c-4)}}}}\\&=\displaystyle \frac{{(c-6)}}{{(c+6)}}\end{align*}
3. .
\scriptsize \begin{align*}&\displaystyle \frac{{16{{a}^{2}}-24a+9}}{{4{{a}^{2}}+17a-15}}\div \displaystyle \frac{{16{{a}^{2}}-9}}{{4{{a}^{2}}+11a+6}}\\&=\displaystyle \frac{{16{{a}^{2}}-24a+9}}{{4{{a}^{2}}+17a-15}}\times \displaystyle \frac{{4{{a}^{2}}+11a+6}}{{16{{a}^{2}}-9}}\\&=\displaystyle \frac{{\bcancel{{(4a-3)}}\bcancel{{(4a-3)}}}}{{\bcancel{{(4a-3)}}(a+5)}}\times \displaystyle \frac{{\bcancel{{(4a+3)}}(a+2)}}{{\bcancel{{(4a+3)}}\bcancel{{(4a-3)}}}}\\&=\displaystyle \frac{{(a+2)}}{{(a+5)}}\end{align*}
4. .
\scriptsize \begin{align*}&\displaystyle \frac{{2{{x}^{2}}+x}}{{2x-1}}\times \displaystyle \frac{{4{{x}^{2}}-1}}{{{{x}^{2}}}}\div \displaystyle \frac{{2x+1}}{x}\\&=\displaystyle \frac{{2{{x}^{2}}+x}}{{2x-1}}\times \displaystyle \frac{{4{{x}^{2}}-1}}{{{{x}^{2}}}}\times \displaystyle \frac{2}{{2x+1}}\\&=\displaystyle \frac{{x\bcancel{{(2x+1)}}}}{{\bcancel{{(2x-1)}}}}\times \displaystyle \frac{{(2x+1)\bcancel{{(2x-1)}}}}{{{{x}^{2}}}}\times \displaystyle \frac{2}{{\bcancel{{(2x+1)}}}}\\&=\displaystyle \frac{{2x}}{{{{x}^{2}}}}\\&=\displaystyle \frac{2}{x}\end{align*}
5. .
\scriptsize \begin{align*}&\displaystyle \frac{{3{{x}^{2}}-21x}}{{18{{x}^{2}}+24x}}\div \displaystyle \frac{{x-7}}{{4x+5}}\cdot \displaystyle \frac{{18{{x}^{3}}-32x}}{{4{{x}^{2}}+x-5}}\div \displaystyle \frac{{3x-4}}{{x-1}}\\&=\left( {\displaystyle \frac{{3{{x}^{2}}-21x}}{{18{{x}^{2}}+24x}}\div \displaystyle \frac{{x-7}}{{4x+5}}} \right)\cdot \left( {\displaystyle \frac{{18{{x}^{3}}-32x}}{{4{{x}^{2}}+x-5}}\div \displaystyle \frac{{3x-4}}{{x-1}}} \right)&&\text{Placing the fractions being divided in brackets makes things clearer}\\&=\left( {\displaystyle \frac{{3{{x}^{2}}-21x}}{{18{{x}^{2}}+24x}}\times \displaystyle \frac{{4x+5}}{{x-7}}} \right)\cdot \left( {\displaystyle \frac{{18{{x}^{3}}-32x}}{{4{{x}^{2}}+x-5}}\times \displaystyle \frac{{x-1}}{{3x-4}}} \right)\\&=\displaystyle \frac{{3{{x}^{2}}-21x}}{{18{{x}^{2}}+24x}}\times \displaystyle \frac{{4x+5}}{{x-7}}\cdot \displaystyle \frac{{2x(9{{x}^{2}}-16)}}{{4{{x}^{2}}+x-5}}\times \displaystyle \frac{{x-1}}{{3x-4}}\\&=\displaystyle \frac{{3x\bcancel{{(x-7)}}}}{{6x\bcancel{{(3x+4)}}}}\times \displaystyle \frac{{\bcancel{{(4x+5)}}}}{{\bcancel{{(x-7)}}}}\cdot \displaystyle \frac{{2x\bcancel{{(3x+4)}}(3x-4)}}{{\bcancel{{(4x+5)}}(x-1)}}\times \displaystyle \frac{{(x-1)}}{{(3x-4)}}\\&=\displaystyle \frac{{6{{x}^{2}}}}{{6x}}\\&=x\end{align*}

Back to Unit 1: Assessment