Functions and Algebra: Manipulate and simplify algebraic expressions

# Unit 3: Simplification of algebraic fractions

Natashia Bearam-Edmunds

### Unit outcomes

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

• Simplify algebraic fractions with monomial denominators.

## What you should know

Before you start this unit, make sure you can:

• Simplify algebraic expressions and factorise. To revise this, go over Units 1 and 2 of this subject outcome.
• Add, subtract, multiply and divide fractions.

Try this short self-assessment to make sure you are ready for this unit.

• Calculate (show all working):
1. $\scriptsize \displaystyle \frac{3}{{20}}+\displaystyle \frac{2}{5}$
2. $\scriptsize \displaystyle \frac{5}{8}+\displaystyle \frac{2}{5}$
3. $\scriptsize \displaystyle \frac{{13}}{{15}}-\displaystyle \frac{2}{5}$
4. $\scriptsize \displaystyle \frac{{12}}{{27}}\div \displaystyle \frac{1}{9}$
5. $\scriptsize \displaystyle \frac{2}{3}\left(\displaystyle \frac{3}{4}+\displaystyle \frac{7}{{10}}\right)$
• Solutions
1. .
$\scriptsize \begin{array}{l}\displaystyle \frac{3}{{20}}+\displaystyle \frac{2}{5}&=\displaystyle \frac{3}{{20}}+\displaystyle \frac{{2\times 4}}{{5\times 4}}\quad (\text{the LCD is }20)\\&=\displaystyle \frac{{3+8}}{{20}}\\&=\displaystyle \frac{{11}}{{20}}\end{array}$
2. .
$\scriptsize \begin{array}{l}\displaystyle \frac{5}{8}+\displaystyle \frac{2}{5}&=\displaystyle \frac{{(5\times 5)+(2\times 8)}}{{40}}\quad \text{(the LCD is }40\text{)}\\&=\displaystyle \frac{{25+16}}{{40}}\\&=\displaystyle \frac{{41}}{{40}}\end{array}$
3. .
$\scriptsize \begin{array}{l}\displaystyle \frac{{13}}{{15}}-\displaystyle \frac{2}{5}&=\displaystyle \frac{{13-2\times 3}}{{15}}\\&=\displaystyle \frac{7}{{15}}\end{array}$
4. .
$\scriptsize \begin{array}{l}\displaystyle \frac{{12}}{{27}}\div \displaystyle \frac{1}{9}&=\displaystyle \frac{{12}}{{27}}\times \displaystyle \frac{9}{1}\\&=\displaystyle \frac{{12}}{3}\\&=4\end{array}$
5. .
$\scriptsize \begin{array}{l}\displaystyle \frac{2}{3}\left(\displaystyle \frac{3}{4}+\displaystyle \frac{7}{{10}}\right)&=\left( {\displaystyle \frac{2}{3}\times \displaystyle \frac{3}{4}} \right)+\left( {\displaystyle \frac{2}{3}\times \displaystyle \frac{7}{{10}}} \right)\\&=\displaystyle \frac{2}{4}+\displaystyle \frac{{14}}{{30}}\\&=\displaystyle \frac{1}{2}+\displaystyle \frac{7}{{15}}\\&=\displaystyle \frac{{(1\times 15)+(7\times 2)}}{{30}}\\&=\displaystyle \frac{{15+14}}{{30}}\\&=\displaystyle \frac{{29}}{{30}}\end{array}$

## Introduction

So far you would have worked with numerical fractions similar to the ones in the short assessment at the beginning of this unit. An algebraic fraction simply means there are variables in the numerator or the denominator of a fraction. We can apply the properties of numerical fractions to algebraic fractions.

A rational expression has a polynomial in the numerator and denominator. For example $\scriptsize \displaystyle \frac{{{{x}^{2}}+2x}}{{x+2}}$ is a rational expression with a binomial in the numerator and a binomial in the denominator. In this unit, the algebraic fractions you will learn to simplify will only have monomials (single terms) in the denominator. $\scriptsize \displaystyle \frac{1}{x},\displaystyle \frac{{5x}}{y}$ and $\scriptsize \displaystyle \frac{3}{{xy}}$ are examples of algebraic fractions with monomial denominators.

## Adding and subtracting simple algebraic fractions

Remember the following techniques for working with numerical fractions.

• $\scriptsize \displaystyle \frac{a}{b}+\displaystyle \frac{c}{b}=\displaystyle \frac{{a+c}}{b}\text{ (}b\ne 0)$
• $\scriptsize \displaystyle \frac{a}{b}-\displaystyle \frac{c}{b}=\displaystyle \frac{{a-c}}{b}\text{ (}b\ne 0)$
• $\scriptsize \displaystyle \frac{a}{b}\times \displaystyle \frac{c}{d}=\displaystyle \frac{{ac}}{{bd}}\text{ (}b\ne 0;d\ne 0)$
• $\scriptsize \displaystyle \frac{a}{b}\div \displaystyle \frac{c}{d}=\displaystyle \frac{a}{b}\times \displaystyle \frac{d}{c}\text{ }(b\ne 0;d\ne 0;c\ne 0)$

Let’s see how we apply these to algebraic fractions.

### Example 3.1

1. $\scriptsize \displaystyle \frac{5}{x}+\displaystyle \frac{3}{y}$
2. $\scriptsize \displaystyle \frac{2}{a}+\displaystyle \frac{a}{3}$
3. $\scriptsize \displaystyle \frac{6}{y}-\displaystyle \frac{5}{{xy}}$

Solutions

1. Just as we do with numerical fractions, we must find the lowest common denominator (LCD) to add fractions with different denominators. The LCD of $\scriptsize \displaystyle \frac{5}{x}+\displaystyle \frac{3}{y}$ is the product of the denominators $\scriptsize x\cdot y$. We then multiply each expression by the appropriate form of $\scriptsize 1$ to get $\scriptsize xy$ as the denominator for each fraction.
.
\scriptsize \begin{align} \displaystyle \frac{5}{x}+\displaystyle \frac{3}{y} & =\displaystyle \frac{5}{x}\cdot \displaystyle \frac{y}{y}+\displaystyle \frac{3}{y}\cdot \displaystyle \frac{x}{x} \\ & =\displaystyle \frac{5y}{xy}+\displaystyle \frac{3x}{xy} && \text{Now that the expressions have the same }\\ &&& \text{denominator, we simply add the numerators}\text{.} \\ & =\displaystyle \frac{5y+3x}{xy} \end{align}
.
Note: Multiplying by $\scriptsize \displaystyle \frac{y}{y}$ or $\scriptsize \displaystyle \frac{x}{x}$ does not change the value of the original expression because any number divided by itself is $\scriptsize 1$ and multiplying an expression by $\scriptsize 1$ gives the original expression.
2. We have to rewrite the fractions so they share a common denominator before we are able to add. The LCD of this algebraic fraction is $\scriptsize 3a$.
$\scriptsize \begin{array}{l}\displaystyle \frac{2}{a}+\displaystyle \frac{a}{3}&=\displaystyle \frac{2}{a}\cdot \displaystyle \frac{3}{3}+\displaystyle \frac{a}{3}\cdot \displaystyle \frac{a}{a}\\&=\displaystyle \frac{6}{{3a}}+\displaystyle \frac{{{{a}^{2}}}}{{3a}}\\&=\displaystyle \frac{{6+{{a}^{2}}}}{{3a}}\end{array}$
3. .
$\scriptsize \begin{array}{l}\displaystyle \frac{6}{y}-\displaystyle \frac{5}{{xy}}&=\displaystyle \frac{6}{y}\cdot \displaystyle \frac{x}{x}-\displaystyle \frac{5}{{xy}}\\&=\displaystyle \frac{{6x}}{{xy}}-\displaystyle \frac{5}{{xy}}\\&=\displaystyle \frac{{6x-5}}{{xy}}\end{array}$

### Exercise 3.1

Simplify:

1. $\scriptsize \displaystyle \frac{a}{4}-\displaystyle \frac{{a-2}}{3}$
2. $\scriptsize \displaystyle \frac{5}{a}+\displaystyle \frac{3}{{ab}}$
3. $\scriptsize \displaystyle \frac{7}{{3x}}-\displaystyle \frac{1}{{2x}}$

The full solutions are at the end of the unit.

## Multiplying and dividing simple algebraic fractions

Multiplication and division of algebraic fractions works in the same way as division and multiplication of other fractions.

Remember that $\scriptsize \displaystyle \frac{2}{x}\times \displaystyle \frac{x}{3}=\displaystyle \frac{{2\cancel{x}}}{{3\cancel{x}}}=\displaystyle \frac{2}{3}$ by cancelling the common factors. Another way of looking at this is to realise that$\scriptsize \displaystyle \frac{x}{x}=1$.

To simplify $\scriptsize \displaystyle \frac{1}{x}\div \displaystyle \frac{3}{x}$ multiply the first expression by the reciprocal of the second.

So we get, $\scriptsize \displaystyle \frac{1}{x}\times \displaystyle \frac{x}{3}$. Once the division expression has been rewritten as a multiplication expression, we can multiply as we did before.

$\scriptsize \begin{array}{l}\displaystyle \frac{1}{x}\times \displaystyle \frac{x}{3}&=\displaystyle \frac{{\cancel{x}}}{{3\cancel{x}}}\\&=\displaystyle \frac{1}{3}\end{array}$

Let’s have a look at a few more examples.

### Example 3.2

Simplify:

1. $\scriptsize \displaystyle \frac{1}{b}\times \displaystyle \frac{2}{b}+\displaystyle \frac{3}{{{{b}^{2}}}}$
2. $\scriptsize \displaystyle \frac{{2y}}{{ab}}\div \displaystyle \frac{4}{a}$
3. $\scriptsize \displaystyle \frac{{y+\displaystyle \frac{1}{x}}}{{\displaystyle \frac{x}{y}}}$

Solutions

1. Remember the order of operations of BODMAS you must multiply before you add.
$\scriptsize \begin{array}{l}\displaystyle \frac{1}{b}\times \displaystyle \frac{2}{b}+\displaystyle \frac{3}{{{{b}^{2}}}}&=\displaystyle \frac{{1\times 2}}{{b\times b}}+\displaystyle \frac{3}{{{{b}^{2}}}}\\&=\displaystyle \frac{2}{{{{b}^{2}}}}+\displaystyle \frac{3}{{{{b}^{2}}}}\\&=\displaystyle \frac{5}{{{{b}^{2}}}}\end{array}$
2. When we divide fractions, we tip the second expression around (find the reciprocal) and multiply.
$\scriptsize \begin{array}{l}\displaystyle \frac{{2y}}{{ab}}\div \displaystyle \frac{4}{a}&=\displaystyle \frac{{\cancel{2}y}}{{\cancel{a}b}}\times \displaystyle \frac{{\cancel{a}}}{{2\cancel{4}}}\text{ Now divide the common factors of }a\text{ and 2}\\&=\displaystyle \frac{y}{{2b}}\end{array}$
3. Begin by combining the expressions in the numerator into one expression.
$\scriptsize \begin{array}{l}y+\displaystyle \frac{1}{x}&=y\cdot \displaystyle \frac{x}{x}+\displaystyle \frac{1}{x}\\&=\displaystyle \frac{{yx}}{x}+\displaystyle \frac{1}{x}\\&=\displaystyle \frac{{xy+1}}{x}\end{array}$
Now the numerator is a single rational expression and the denominator is a single rational expression.
$\scriptsize \displaystyle \frac{{\displaystyle \frac{{xy+1}}{x}}}{{\displaystyle \frac{x}{y}}}$
We can rewrite the above expression as division, and then multiplication.
$\scriptsize \begin{array}{l}\displaystyle \frac{{xy+1}}{x}\div \displaystyle \frac{x}{y}&=\displaystyle \frac{{xy+1}}{x}\times \displaystyle \frac{y}{x}\\&=\displaystyle \frac{{y(xy+1)}}{{{{x}^{2}}}}\end{array}$

As you can see from the example, algebraic fractions can be simplified by cancelling common factors in the numerator and denominator.

### Exercise 3.2

Simplify:

1. $\scriptsize \displaystyle \frac{{5x}}{{6{{y}^{2}}}}\times \displaystyle \frac{{3y}}{{15x}}$
2. $\scriptsize \displaystyle \frac{{x+2}}{{4{{z}^{2}}}}\div \displaystyle \frac{1}{{4{{z}^{3}}}}$
3. $\scriptsize \displaystyle \frac{{\displaystyle \frac{x}{y}-\displaystyle \frac{y}{x}}}{y}$

The full solutions are at the end of the unit.

## Summary

In this unit you have learnt the following:

• How to define an algebraic fraction.
• How to simplify simple algebraic fractions with monomial denominators.
• How to add and subtract algebraic fractions with monomial denominators.
• How to multiply and divide algebraic fractions with monomial denominators.
• How to simplify complex fractions with monomial denominators.

# Assessment

#### Suggested time to complete: 15 minutes

1. Simplify:
1. $\scriptsize \displaystyle \frac{x}{y}+\displaystyle \frac{2}{x}$
2. $\scriptsize \displaystyle \frac{a}{{2b}}-\displaystyle \frac{{2b}}{{9a}}$
3. $\scriptsize \displaystyle \frac{1}{2}-\displaystyle \frac{b}{{3a}}$
$\scriptsize \displaystyle \frac{{\displaystyle \frac{a}{{2b}}-\displaystyle \frac{{2b}}{{9a}}}}{{\displaystyle \frac{1}{2}-\displaystyle \frac{b}{{3a}}}}$

The full solutions are at the end of the unit.

# Unit 3: Solutions

### Exercise 3.1

1. .
$\scriptsize \begin{array}{l}\displaystyle \frac{a}{4}-\displaystyle \frac{{a-2}}{3}&=\displaystyle \frac{{a\times 3-(a-2)\times 4}}{{12}}\\&=\displaystyle \frac{{3a-4a+8}}{{12}}\\&=\displaystyle \frac{{-a+8}}{{12}}\end{array}$
2. .
$\scriptsize \begin{array}{l}\displaystyle \frac{5}{a}+\displaystyle \frac{3}{{ab}}&=\displaystyle \frac{{5\cdot b+3}}{{ab}}\text{ LCD is }ab\text{ so only the }5\text{ is multiplied by }b\\&=\displaystyle \frac{{5b+3}}{{ab}}\end{array}$
3. .
$\scriptsize \begin{array}{l}\displaystyle \frac{7}{{3x}}-\displaystyle \frac{1}{{2x}}&=\displaystyle \frac{7}{{3x}}\cdot \displaystyle \frac{2}{2}-\displaystyle \frac{1}{{2x}}\cdot \displaystyle \frac{3}{3}\quad \text{LCD is }6x\text{. }3x\times 2=6x\text{ and }2x\times 3&=6x\\&=\displaystyle \frac{{14}}{{6x}}-\displaystyle \frac{3}{{6x}}\\&=\displaystyle \frac{{14-3}}{{6x}}\\&=\displaystyle \frac{{11}}{{6x}}\end{array}$

Back to Exercise 3.1

Exercise 3.2

1. .
$\scriptsize \begin{array}{l}\displaystyle \frac{{5x}}{{6{{y}^{2}}}}\times \displaystyle \frac{{3y}}{{15x}}&=\displaystyle \frac{{\cancel{5}\cancel{x}}}{{2y}}\times \displaystyle \frac{{\cancel{3}\cancel{y}}}{{3\cancel{x}}}\\&=\displaystyle \frac{1}{{6y}}\end{array}$
2. .
$\scriptsize \begin{array}{l}\displaystyle \frac{{x+2}}{{4{{z}^{2}}}}\div \displaystyle \frac{1}{{4{{z}^{3}}}}&=\displaystyle \frac{{x+2}}{{4{{z}^{2}}}}\times \displaystyle \frac{{4{{z}^{3}}}}{1}\\&=\displaystyle \frac{{x+2}}{{\cancel{4}{{{\cancel{z}}}^{{\cancel{2}}}}}}\times \displaystyle \frac{{\cancel{4}{{z}^{{\cancel{3}}}}}}{1}\\&=z(x+2)\end{array}$
3. .
$\scriptsize \displaystyle \frac{{\displaystyle \frac{x}{y}-\displaystyle \frac{y}{x}}}{y}=\displaystyle \frac{x}{y}-\displaystyle \frac{y}{x}\div \displaystyle \frac{1}{y}$
Work with the numerator first.
$\scriptsize \begin{array}{l}\displaystyle \frac{x}{y}-\displaystyle \frac{y}{x}&=\displaystyle \frac{{x\cdot x-y\cdot y}}{{xy}}\\&=\displaystyle \frac{{{{x}^{2}}-{{y}^{2}}}}{{xy}}\end{array}$
Then bring back the divisor.
$\scriptsize \begin{array}{l}\displaystyle \frac{{{{x}^{2}}-{{y}^{2}}}}{{xy}}\div \displaystyle \frac{1}{y}&=\displaystyle \frac{{{{x}^{2}}-{{y}^{2}}}}{{x\cancel{y}}}\times \displaystyle \frac{{\cancel{y}}}{1}\\&=\displaystyle \frac{{{{x}^{2}}-{{y}^{2}}}}{x}\end{array}$

Back to Exercise 3.2

### Assessment

Simplify:

1. .
$\scriptsize \begin{array}{l}\displaystyle \frac{x}{y}+\displaystyle \frac{2}{x}&=\displaystyle \frac{{x\cdot x+2\cdot y}}{{xy}}\\&=\displaystyle \frac{{{{x}^{2}}+2y}}{{xy}}\end{array}$
1. .
$\scriptsize \begin{array}{l}\displaystyle \frac{a}{{2b}}-\displaystyle \frac{{2b}}{{9a}}&=\displaystyle \frac{{9\cdot a\cdot a}}{{18ab}}-\displaystyle \frac{{2b\cdot (2b)}}{{18ab}}\\&=\displaystyle \frac{{9{{a}^{2}}-4{{b}^{2}}}}{{18ab}}\end{array}$
2. $\scriptsize \displaystyle \frac{1}{2}-\displaystyle \frac{b}{{3a}}=\displaystyle \frac{{3a-2b}}{{6a}}$
2. From b) and c) we get:
$\scriptsize \displaystyle \frac{{\displaystyle \frac{a}{{2b}}-\displaystyle \frac{{2b}}{{9a}}}}{{\displaystyle \frac{3}{2}-\displaystyle \frac{b}{{3a}}}}=\displaystyle \frac{{9{{a}^{2}}-4{{b}^{2}}}}{{18ab}}\div \displaystyle \frac{{3a-2b}}{{6a}}$
Next, we multiply by the reciprocal:
$\scriptsize \displaystyle \frac{{9{{a}^{2}}-4{{b}^{2}}}}{{18ab}}\times \displaystyle \frac{{6a}}{{3a-2b}}$
To simplify even further we must factorise the numerator of the first expression using a difference of two squares.
$\scriptsize \begin{array}{l}\displaystyle \frac{{9{{a}^{2}}-4{{b}^{2}}}}{{18ab}}\times \displaystyle \frac{{6a}}{{3a-2b}}&=\displaystyle \frac{{(3a-2b)(3a+2b)}}{{18ab}}\times \displaystyle \frac{{6a}}{{(3a-2b)}}\\&=\displaystyle \frac{{(3a+2b)}}{{3\text{ }\cancel{1}\cancel{8}\cancel{a}b}}\times \displaystyle \frac{{\cancel{6}\cancel{a}}}{1}\quad \text{Factors of }(3a-2b)\text{ cancel}\\&=\displaystyle \frac{{3a+2b}}{{3b}}\end{array}$

Back to Unit 3: Assessment